# Worksheets For Class 10 Mathematics Quadratic Equation

## Mathematics Quadratic Equation Worksheets for Class 10

We have provided chapter-wise worksheets for class 10 Mathematics Quadratic Equation which the students can download in Pdf format for free. This is the best collection of Mathematics Quadratic Equation standard 10th worksheets with important questions and answers for each grade 10th Mathematics Quadratic Equation chapter so that the students are able to properly practice and gain more marks in Class 10 Mathematics Quadratic Equation class tests and exams.

### Chapter-wise Class 10 Mathematics Quadratic Equation Worksheets Pdf Download

Question. If α, β, be the roots of the equation ax2 + bx + c = 0, then ax2 + bx + c =.
(A) (x – α)(x – β)
(B) a(x – α)(x – β)
(C) a (x – β) (x + α)
(D) a(x + α)(x + β)

C

Question.The number of real roots of the equation (x – 1)2 + (x – 2)2 + (x – 3)2 = 0 is:
(A) 2
(B) 1
(C) 0
(D) 3

C

Question.Determine k such that the quadratic equation x2 + 7(3 +2k) – 2x(1 + 3k) = 0 has equal roots:
(A) 2, 7
(B) 7, 5
(C) 2, -10/9
(D) None of these

C

Question. The condition that the equation ax2 + bx + c = 0 has one positive and other negative root is:
(A) a and c will have same sign
(B) a and c will have opposite signs
(C) b and c will have same sign
(D) b and c will have opposite signs

B

Question. The condition that both roots of the equation ax2 + bx + c = 0 are negative is:
(A) a, b, c are of the same sign
(B) a and b are of opposite signs
(C) b and c are of opposite signs
(D) the absolute term is zero

A

Question. If one root of 5x2 + 13x + k = 0 is reciprocal of the other, then the value of k is.
(A) 5
(B) 7
(C) 4
(D) None of these

A

Question. If the equation x2-bx / ax-c = m-1/m+1 has roots equal in magnitude but opposite in sign, then m is equal to.
(A) a+b /a -b
(B)  a-b/a+b
(C) Both (A) & (B)
(D) None of these

B

Question. The set of values of p for which the roots of the equation 3x2 + 2x + (p – 1)p = 0 are of opposite sign, is:
(A) (0, 1)
(B) (-1,1)
(C) (-2,2)
(D) None of these

A

Question. Find the roots of the equation f(x) = (b – c) x2 + (c – a) x + (a – b) = 0
(A)  a-b /b-c and 1
(B) a+b /b+c and 1
(C) a-b /b+c and 1
(D) None of these

A

Question. The real values of a for which the quadratic equation 2x2 – (a3 + 8a – 1)x + a2 – 4a = 0 possesses roots of opposite signs are given by:
(A) a > 6
(B) a > 9
(C) 0 < a < 4
(D) a < 0

C

Question. Given that (x + 1) is a factor of x2 + ax + b and x2 + cx – d, then
(A) a + d = b + c
(B) a = b + c + d
(C) a + c = b – d
(D) None of these

B

Question. The zeroes of x2 – bx + c are each decreased by 2 The resulting polynomial is x2 – 2x + 1 Then
(A) b = 6, c = 9
(B) b = 6, c = 3
(C) b = 3, c = 6
(D) b = 9, c = 6

A

Question. If the zeroes of x2 + Px + t are two consecutive even numbers find the relation between P and t
(A) P2 – 4t + 4 = 0
(B) 4t – P2 + 4 = 0
(C) – 4t2 – 4 – P2 = 0
(D) None of these

B

Question. One zero of x2 – bx + C is the kth power of the other zero, then C 1/k+1+Ck/k+1 is equal to
(A) – b
(B) C
(C) – C
(D) b

D

Question. α, β, γ, δ are zeroes of x4 + 5x3 + 5x2 + 5x – 6, then find the value of 1/α + 1/β + 1/γ +1/δ
(A) 5/6
(B) -6/5
(C) -5/6
(D) None of these

A

Question. The number of real solution of the equation 23×2-7x + 4 = 1 is:
(A) 0
(B) 4
(C) 2
(D) Infinitely many

C

Question. The maximum value of – 3x2 + 4x – 5 is at x =
(A) 2/3
(B) 1/3
(C) -33/9
(D) None of these

A

Question. Given that 7 – 3i is a zero of x2 + px + q, find the value of 3q + 4p
(A) 14
(B) 58
(C) 118
(D) – 14

C

Question. Given that α is a zero of x4 + x2 – 1, find the value of (α6 + 2α4)1000.
(A) 1
(B) 0
(C) Either 0 or 1
(D) None of these

A

VERY SHORT ANSWER TYPE QUESTIONS

Question. If one root of the equation 3x2 + px + 4 = 0 is 2/3 , find the value of p.

p = -8

Question. Determine whether x = – 1/3 and x = 2/3 are the roots of 9x2 – 3x – 2 = 0.

yes

Question. If sum of a whole number and its reciprocal is 10/3 , what is the number?

3

Question. For what value of k, x = a is a solution of the equation x2 – (a + b)x + k = 0 .

k = ab

Question. Find the ratio of sum and product of the roots of the equation 3x2 – 8x + 12 = 0.

2 : 3

Question. Form a quadratic equation whose roots are –2 and 3.

x– x – 6 = 0

Question. If one root of a quadratic equation is 3 – √2/4 , then what is the other root?

3 + √2/4

Question. Form a quadratic equation whose sum of roots is –3 and product of roots is 5.

x2 + 3x + 5 = 0

Question Write the quadratic equation for the following : The sum of two numbers is 27 and their product is 182.

x2 – 27x +182 = 0

Question. The sum of squares of two consecutive positive integers is 365. Express this in the form of an quadratic equation.

x2 + x – 182 = 0

Question. Represent the following in the form of a quadratic equation : The sum of the ages of father and his son is 45 years. Five years ago, the product of their ages (in years) was 124.

x2 – 45x + 324 = 0

Question. Write the discriminant of the equation 6a2x2 – 7abx – 3b2 = 0.

121a2b2

Question. Write the nature of the roots of the equation 3x2 – 4 √3x – 1 = 0.

Real and distinct roots

Question. Check whether x (x + 3) + 8x = (x + 5) (x – 5) is a quadratic equation.

No

Question. For what value of k, the equation 3x2 + kx + 4 = 0 have equal roots?

k = ±4√3

Question. If x = 3/2 is the root of the quadratic equation 2x2 – kx + 3 = 0, then find the value of k.

k = 5

Question. For what value of k, the equation kx2 + 6x + 1 = 0 have real roots?

k ≤ 9

Question. If α and β are the roots of the equation 3x2 – 9x + 7 = 0, then find α + β + αβ.

16/3

Question. What are the roots of the equation 2x2 + 3x – 2 = 0?

-2 , 1/2

Question. Solve the quadratic equation : (x + 3)2 – 16 = 0.

1, –7

Short Answer type Question :

Question. Find the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other.
Solution. Let the roots of the given equation be a and 1/α
α , 1/α = c/a = k/3 ⇒ k = 3

Question. Find the value of k so that the quadratic equation kx(3x – 10) + 25 = 0, has two equal roots.
Solution. kx(3x – 10) + 25 = 0
⇒ 3kx2 – 10kx + 25 = 0
D = (–10k)2 – 4 × 3k × 25
= 100k2 – 300k
For equal roots, D = 0 (1)
⇒ 100k2 – 300k = 0
⇒ 100k(k – 3) = 0
⇒ k = 0 or k = 3
But k π 0,
So, k = 0 (Rejected)
[∴ In quadratic equation, a ≠ 0]
Hence, k = 3

Question. For what values of k, the equation 9x2 + 6kx + 4 = 0 has equal roots?
Solution. 9x2 + 6kx + 4 = 0
(6k)2 – 4 × 9 × 4 = 0 (½)
36k2 = 144
⇒ k2 = 4
k = ±2

Question. Find the value(s) of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots.
Solution. x2 – 4kx + k = 0
Since given equation has equal roots,
∴ D = 0 (1)
16k2 – 4k = 0
⇒ 4k(4k – 1) = 0
⇒ k = 0 and k = 1/4

Question. For what value(s) of ‘a’ quadratic equation 3ax2 – 6x + 1 = 0 has no real roots?
Solution. 3ax2 – 6x + 1 = 0 (½)
(–6)2 – 4(3a)(1) < 0
12a > 36
⇒ a > 3

Question. State whether the quadratic equation 4x2 – 5x + 25/16 = 0 has two distinct real roots or not. Justify your answer.
Solution. No, D = 0

Question. Find the value(s) of k so that the quadratic equation 3x2 – 2kx + 12 = 0 has equal roots.
Solution. 3x2 – 2kx + 12 = 0
Since given equation has equal roots, so
D = 0 (1)
⇒ 4k2 – 144 = 0
⇒ 4(k2 – 36) = 0
⇒ k = ±6
⇒ k = 6 and k = –6

Question. Find the value of p, so that the quadratic equation px(x – 3) + 9 = 0 has equal roots.
Solution. px(x – 3) + 9 = 0
⇒ px2 – 3px + 9 = 0
When roots are equal,
D = b2 – 4ac = 0
9p2 – 36p = 0
⇒ 9p(p – 4) = 0
⇒ p = 0, p = 4
But p π 0
[∴ In quadratic equation, a ≠ 0]
∴ p = 4

Question. Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.
Solution. For equal roots, D = 0
⇒ 9k2 – 36k = 0
⇒ 9k(k – 4) = 0
⇒ k = 0 or k = 4

Question. Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, hence find the nature of its roots.
Solution. –8, no real roots

Question. If x = 3 is one root of the quadratic equation x2 – 2kx– 6 = 0, then find the value of k.
Solution. x = 3 is one root of the equation
∴ 9 – 6 k – 6 = 0
⇒ k = 1/2

Question. Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0 has equal roots. Also find these roots.
Solution.  For equal roots, D = 0
{2(k + 1)}2 – 4(3k + 1) · 1 = 0 (1)
⇒ 4(k2 + 2k + 1) – 12k – 4 = 0
⇒ 4k2 + 8k + 4 – 12k – 4 = 0 (1)
⇒ 4k2 – 4k = 0
⇒ 4k(k – 1) = 0
⇒ k = 0, 1

Question. Find the nature of the roots of the following quadratic equations. If the real roots exist, then also find them.
(a) 4x2 + 12x + 9 = 0 (b) 3x2 + 5x – 7 = 0 (c) 7y2 – 4y + 5 = 0
Solution.

Question. For what value of k does the quadratic equation (k – 5)x2 + 2(k – 5)x + 2 = 0 have equal roots?
Solution. k = 7

Question. Check whether the equation 5x2 – 6x – 2 = 0 has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
Solution. Discriminant = b2 – 4ac
= 36 – 4 × 5 × (–2)
= 76 > 0
So, the given equation has two distinct real roots
5×2 – 6x – 2 = 0 (1)
Multiplying both sides by 5, we get
(5x)2 – 2 × (5x) × 3 = 10
⇒ (5x)2 – 2 × (5x) × 3 + 32 = 10 + 32
⇒ (5x – 3)2 = 19
⇒ 5x – 3 = ± √19

Question. Find the value(s) of k so that the quadratic equation 2x2 + kx + 3 = 0 has equal roots.
Solution. 2x2 + kx + 3 = 0
For equal roots, D = 0
⇒ b2 – 4ac = 0
⇒ k2 – 24 = 0
⇒ k = ± 2√6

Question. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0 has equal roots, find the value of k.
Solution. 3(2)2 + p(2) – 8 = 0
⇒ 12 + 2p – 8 = 0
⇒ p = – 2 …(i)(1)
So, equation becomes
4x2 + 4x + k = 0
For equal roots, D = 0
⇒ (4)2 – 4 × 4 × k = 0
⇒ 16 = 16k
⇒ k = 1

Question. Find the value of p for which the quadratic equation (p + 1)x2 – 6(p + 1)x + 3(p + 9) = 0, p π –1 has equal roots. Hence, find the roots of the equation.
Solution. 3, 3.

Question. For what values of k, the roots of the equation x2 + 4x + k = 0 are real?
Solution. For real roots, D ≥ 0
⇒ b2 – 4ac ≥ 0
⇒ (4)2 – 4 × 1 × k ≥ 0
⇒ 16 – 4k ≥ 0
⇒ 16 ≥ 4k, k ≤ 4

Question. Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k – 1)x + x2 = 0 has equal roots. Hence, find the roots of the equation.
Solution. kx2 + 1 – 2(k – 1)x + x2 = 0
⇒ (k + 1)x2 – 2(k – 1)x + 1 = 0
Q Above equation has equal roots,
So, discriminant, D = 0
⇒ {–2(k – 1)}2 – 4 × (k + 1) × 1 = 0
⇒ 4(k2 – 2k + 1) – 4(k + 1) = 0
⇒ 4k2 – 12k = 0
⇒ 4k(k – 3) = 0
⇒ k = 3 (as k ≠ 0)

Question. Find whether the equation 1/2x – 3 + 1/x – 5 = 1 , x ≠ 3/2 , 5 has real roots. If real roots exist, find them.
Solution. Yes, 8 ±3√2/2

Question. The roots a and b of the quadratic equation x2 – 5x + 3(k – 1) = 0 are such that a – b = 1. Find the value k.
Solution. k = 3

Question. Solve for x: 2(x + 2)/2x -3 – 9(2x -3)/x + 2 = 3; given that x ≠ -2, x ≠ 3/2
Solution.

Question. Find the values of k for which the given equation has real and equal roots: (k + 1)x2 -2(k – 1)x + 1 = 0
Solution. We have, (k+1)x2 – 2(k – 1)x+1 = 0.
a = k + 1, b = -2(k – 1), c = 1.
D = b2 – 4ac =4(k-1)2 – 4(k + 1) =4(k2 -3k)
The given equation will have real and equal roots, if
D = 0 ⇒ 4 (k2 – 3k) = 0 ⇒ k2 – 3k = 0 ⇒ k (k – 3) = 0 ⇒ k = 0, 3

Question. Solve for x: √3x2 + 10x + 7√3 = 0
Solution. We have the following equation,
√3x2 + 10x + 7√3 = 0
Now factorise the equation,

Question. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution. Let Shefali’s marks in Mathematics = x
Let Shefali’s marks in English = 30 − x
If, she had got 2 marks more in Mathematics, her marks would be = x + 2
If, she had got 3 marks less in English, her marks in English would be = 30 – x − 3 = 27
− x
According to given condition:
⇒ (x + 2)(27 − x) = 210
⇒ 27x – x2 + 54 – 2x = 210
⇒ x2 – 25x + 156 = 0
Comparing quadratic equation x2 – 25x + 156 = 0 with general form
ax2 + bx + c = 0
We get a = 1, b = −25 and c = 156

Therefore, Shefali’s marks in Mathematics = 13 or 12
Shefali’s marks in English = 30 – x = 30 – 13 = 17
Or Shefali’s marks in English = 30 – x = 30 – 12 = 18
Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).

Question. If x = -1/2 is a solution of the quadratic equation 3×2 + 2kx + 3 = 0, find the value of k.
Solution.

Question. Check, whether the quadratic equation have real roots and if so, then find the roots of equation. 6x2 + x – 2 = 0
Solution. The given equation is 6x2 + x – 2 = 0
Here, a = 6, b = 1 and, c = – 2

Question. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2– 2px + k = 0 has equal roots, find k.
Solution. Given, 2 is a root of the equation, 3x2 + px – 8 = 0
Putting x = 2 in 3x2 + px – 8 = 0
12 + 2p – 8 = 0
or, p = – 2

Question. Solve the quadratic equations by factorization method: x2 – 9 = 0
Solution. We have,
x2 – 9 = 0
⇒ (x – 3)(x + 3) = 0
⇒ x – 3 = 0 or, x + 3 = 0
x = 3 or, x = -3 ⇒ x = ± 3
Thus, x = 3 and x = – 3 are roots of the given equation.

Question. If p, q, r and s are real numbers such that pr = 2(q + s), then show that at least one of the equations x2 + px + q = 0 and x2 + rx + s = 0 has real roots.
Solution. Given quadratic equations are;
x2 + px + q = 0 —(i)
and, x2 + rx + s = 0 ……(ii)
Also given ; pr = 2(q + s)……..(iii)
Let D1 and D2 be the discriminant of quadratic equations (i) and (ii) respectively.
Then,
D1 = p2 – 4q and D2 = r2 – 4s
⇒ D1+ D2 = p2 – 4q + r2 – 4s = (p2 + r2) – 4(q + s)
Now, Since sum of both D2 & D1 is greater than or equal to 0. Hence, both can’t be
negative.
⇒ At least one of D1and D2 is greater than or equal to zero
Case 1. If D1 ≥ 0, equation (i) has real roots.
Case 2.If D2 ≥ 0, equation (ii) has real roots.
Case 3. If D1 & D2 both ≥ 0, then equation (i) & (ii) both have equal roots.
Clearly, from case 1,2 & 3 at least one given quadratic equations has equal roots.

Question. Write the discriminant of the given quadratic equation x2 + x – 12 = 0
Solution. The given quadratic equation is x2 + x – 12 = 0
here a=1, b=1, c=-12

Hence, the discriminant is 49.

Question. Check whether the given equation is quadratic equation: (x-3) (2x + 1) = x(x + 5)
Solution. The given equation is (x – 3) (2x +1) = x (x+5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ x2 – 10x – 3 =
It is in the form of ax2 + bx + c = 0, a≠0
∴ the given equation is a quadratic equation.

Question. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.
Solution. Given, speed of boat in still water = 8 Km/hr. Let the speed of the stream be x km/hr.
Then,
Speed of boat in downstream = (8 + x) km/hr
Speed of boat in upstream = (8 – x) km/hr
We know that time taken to cover ‘d’ km with speed ‘s’ km/hr is d/s
So,Time taken by the boat to go 15 km upstream = 15/8 – x  hours.
&, Time taken by the boat to 22 km downstream = 22/8 = x hours.
It is given that the total time taken by boat to go 15 km upstream & 22 km downstream is 5 hours.

⇒ -7x + 296 = 5(64 – x2)
⇒ -7x + 296 = 320 – 5x2
⇒ 5x2 – 7x + 296 – 320 = 0
⇒ 5x2 – 7x – 24 = 0
⇒ 5x2 – 15x + 8x – 24 = 0
⇒ 5x(x – 3) + 8(x – 3) = 0
⇒ (5x + 8)(x- 3) = 0
⇒ x – 3 = 0 [ Speed can not be negative ∴ 5x + 8 ≠ 0]
⇒ x = 3
Hence, the speed of the stream is 3 km/hr.

Question. Find the values of p for which the quadratic equation 4x2 + px + 3 = 0 has equal roots.
Solution.

Question. A train travelling at a uniform speed for 360 km,would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train.
Solution. Given that a train travelling at a uniform speed for 360 km
Let the original speed of the train be x km/hr

Either x = – 50 or x = 45
As speed cannot be negative
∴ Original speed of train = 45 km/hr.

Question. Form a quadratic equation whose roots are -3 and 4.
Solution. We have, x = 4 and x = -3.
Then,
x – 4 = 0 and x + 3 = 0
⇒ (x – 4)(x + 3) = 0
⇒ x2 + 3x – 4x – 12 = 0
⇒ x2 – x – 12 = 0
This is the required quadratic equation

Question. Check whether the given equation is quadratic equation: (x +1)2= 2 (x –3)
Solution. The given equation is (x+1)2 = 2 (x-3)
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 7 = 0
⇒ x2 + 0.x + 7 = 0
Which is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

Question. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2. Find the length and breadth of the pond
Solution.

Let width of the pond be x m. Then,
The length of pond = (50 – 2x)m and the breadth of pond = (40 – 2x)m
Area of grass around the pond = 1184 m2
⇒ Area of Lawn – Area of Pond = 1184
⇒ 50 x 40 – (50 – 2x)(40 – 2x) = 1184
⇒ 2000 – (2000 – 100x – 80x + 4x2) – 1184 = 0
⇒ 2000 – (2000 – 180x + 4x2) – 1184 = 0
⇒ 2000 – 2000 + 180x – 4x2 – 1184 = 0
⇒ 4x2 – 180x + 1184 = 0
⇒ 4(x2 – 45x + 296) = 0
⇒ x2 – 45x + 296 = 0
Factorise now,
⇒ x2 – 37x – 8x + 296 = 0
⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0
⇒ x – 37 = 0 or x – 8 = 0
⇒ x = 37 or x = 8
When x = 37, then
The length of pond = 50 – 2 x 37
= 50 – 74
= -24 m (Length cannot be negative)
When x = 8, then
The length of pond = 50 – 2x
= 50 – 2 x 8
= 50 – 16 = 34 m
And the breadth of the pond
= 40 – 2x
= 40 – 2 x 8
= 40 – 16 = 24 m
Therefore, the length and breadth of the pond are 34 m and 24 m respectively.

Question. Find discriminant of the quadratic equation: 5x2 + 5x + 6 = 0.
Solution. Given equation is 5x2 + 5x + 6 = 0
Here a = 5, b = 5, c = 6
D = b2 – 4ac = (5)2 -4 x 5 x 6 = -95

Question. Two numbers differ by 3 and their product is 504. Find the numbers.
Solution. Sol : Let the required number be x and x + 3.
Then, according to given question we have,
x (x + 3) = 504
⇒ x2 + 3x = 504
⇒ x2 + 3x – 504 = 0
⇒ x2 + 24x – 21x – 504 = 0
⇒ x(x + 24) – 21(x + 24) = 0
⇒ (x + 24)(x – 21) = 0
⇒ x + 24 = 0 or x – 21 = 0
⇒ x = – 24 or x = 21
Case I: When x = -24
∴ x + 3 = -24 + 3 = -21
Case II: When x = 21
∴ x + 3 = 21 + 3 = 24
Hence, the numbers are -21, -24 or 21, 24.

Question. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Solution. Let the smaller number be x and the larger number be y.
Also, Square of the larger number( y2 ) =18x
According to question,
x2 + y2 = 208
⇒ x2 + 18x = 208
⇒ x2 + 18x – 208 = 0
⇒ x2 + 26x – 8x – 208 = 0
⇒ (x + 26) (x – 8) = 0 ⇒ x = 8, x = -26
But, the numbers are positive. Therefore, x =8
Square of the larger number = 18x = 18 x 8 = 144
Therefore, larger number =√144 = 12
Hence, the numbers are 8 and 12.

Question. Solve the quadratic equation by factorization: 3x2 – 2√6x + 2 = 0
Solution.

Question. State whether the following equation is quadratic equation in x? 2x2 + 5/2 x – √3 = 0
Solution.

Question. A man travels a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km an hour, the journey would have taken two hours less. Find the original speed of the train.
Solution. Let the original speed of the train be x km an hour
Then, the total speed taken by the train to travel a distance of 300 km at a uniform
speed of x km an hour 300/x = hours
Increased speed of the train = ( x + 5) km an hour
Time taken by the train to travel a distance of 300 km at the increased speed = 300/x + 5 hours

Comparing with ax2 + bx +c = 0
a = 1, b = 5, c = -750

x = – 30 is inadmissible as x is the speed of the train and speed cannot be negative.
∴ x = 25
Hence, the original speed of the train is 25 km an hour.

Question. If -2 is a root of the equation 3x2 + 5x + 2k = 0, then find the value of k.
Solution.

Question. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously, fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.
Solution. Let the number of hours required by the second pipe alone to fill the pool be x hrs.
Then, the first and third pipe takes (x+5)hrs, (x-4) hrs respectively to fill the pool.
So, the parts of the pool filled by the first, second and third pipes in one hour are

Let the time taken by first and second pipes to fill the pool simultaneously be t hrs.
Then the third pipe also takes the same time to fill the pool

Since time taken cannot be negative. So x =10
Hence, the time required by the first, second and the third pipes to fill the pool
individually are 15 hrs, 10 hrs and 6 hrs respectively.

Question. The sum of first 2 even natural numbers is given by the relation x = n (n + 1). Find n, if the sum is 420.
Solution. n(n + 1) = 420 …Given

Question. Find two consecutive numbers whose squares have the sum 85.
Solution. Let the two consecutive numbers be x and x + 1.
According to question,
x2 + (x + 1)2 = 85

Hence, numbers are 6 and 7 or -7 and -6

Question. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is km/h less than that of the fast train, find the speeds of the two trains.
Solution. Let the speed of the slow train be x km/hr
Then, the speed of the fast train = (x+10) km/hr

Either x = -50 or x = 40
But the speed of the train cannot be negative. So, x= 40
Hence, the speed of the two trains are 40km/hr and 50km/hr respectively.

Question. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
Solution.

Let base = x
Altitude = y
Hypotenuse = h

Question. The difference of two numbers is 4. If the difference of their reciprocals is 4/21 , then find the two numbers.
Solution. Let first number be x.
Then, second number = x + 4
According to the question,

4x2 + 4x = 84
⇒ 4x2 + 16x – 84 = 0
⇒ 4(x2 +4 x – 21) = 0
⇒ (x2 +4 x – 21) = 0
⇒(x +7) (x -3) = 0
⇒ x +7 = 0 or x – 3 = 0
⇒ x = -7 or x = 3
Therefore, the two numbers are 3 and 7 or -7 and -3

Question. Show that the equation x2 + 6x + 6 = 0 has real roots and solve it.
Solution. We have the following equation,
x2 + 6x + 6 = 0 where
a = 1, b = 6 and c = 6.
∴ D = (b2 – 4ac) = (36 – 4 x 1 x 6) = 12 > 0
So, the given equation has real roots.

Question. The product of Tanvy’s age (in years) 5 years ago and her age 8 years later is 30. Find her present age.
Solution. Let the present age of tanvy be x years
Tanvy’s age five years ago = (x – 5)
Tanvy’s age eight years later = (x + 8)
According to question,
⇒ (x – 5)(x + 8) = 30
⇒ x2 + 8x – 5x – 40 = 30
⇒ x2 + 3x – 40 – 30 = 0
⇒ x2 + 3x – 70 = 0
⇒ x2 + 10x – 7x – 70 = 0
⇒ x(x + 10) – 7(x + 10) = 0
⇒ x + 10 = 0 or x – 7 = 0
⇒ x = -10 or x = 7
⇒ x = 7 ( age cannot be negative)
Therefore, the present age of tanvy is 7 years.

Question. Solve: 4x2 – 12x + 9 = 0.
Solution. We have,
4x – 12x + 9 = 0
Here, 4 x 9 = 36 so to factor the middle term in given equation we have (-6) x (-6) = 36, and (-6) + (-6) = -12.
⇒ 4x2 – 6x – 6x + 9 = 0 ⇒ 2x(2x – 3) – 3(2x – 3) = 0
⇒ (2x – 3)(2x – 3) = 0 ⇒ (2x – 3)2 = 0
2x – 3 = 0 ⇒ x = 3/2
Hence, x = 3/2 is the repeated root of the given equation.

Question. If –4 is a root of the quadratic equation x2 + px – 4 = 0 and the quadratic equation x2 + px + k = 0 has equal roots. Find the value of k.
Solution. We have, x2 + px – 4 = 0
-4 is the root of the given equation
Substitute x = – 4 in the given equation, we get
(-4)2 + p (-4) -4 =0
⇒ 16 – 4p – 4 = 0
⇒ 4p = 12 or p = 3
The equation becomes x2 + 3x -4 = 0
Substituting the value of p = 3 in the equation x2 + px + k = 0, we get
x2 + 3x + k = 0
Here a = 1, b = 3, c= k
∴ D = b2 – 4ac = (3)2 – 4 (1) (k)
= 9 – 4k
For equal roots, D = 0
⇒ 9 – 4k = 0 or k = 9/4

Question. Solve the quadratic equations by factorization method 4/x – 3 = 5/2x + 3 , x ≠ 0 , -3/2
Solution. Given,

Question. If p, q and r are rational numbers and p q r, then find the roots of the equation (p2 – q2)x2 – (q2 – r2)x + r2 – p2 = 0.
Solution.

Question. If roots of the quadratic equation x2 + 2px + mn = 0 are real and equal, show that the roots of the quadratic equation x2 – 2(m + n)x + (m2 + n2 + 2p2) = 0 are also equal.
Solution.

Question. Some students planned a picnic. The total budget for hiring a bus was Rs. 1440. Later on, eight of these refused to go and instead paid their total share of money towards the fee of one economically weaker student of their class, and thus, the cost for each member who went for picnic, increased by Rs. 30.
i. How many students attended the picnic?
ii. How much money in total was paid towards the fee? Which value is reflected in this question?
Solution. Let x students planned the picnic.
Then, (x – 8) students attended the picnic.
Total bus charges = Rs. 1440

Thus, 24 students planned the picnic.
i. Number of students who attended the picnic = (24 – 8) = 16
ii. Share of 24 students = Rs. 1440
Share of 8 students = Rs. (1440/24 x 8 = Rs. 480
∴ money paid towards the fee = Rs. 480
The value reflected in the given question is ‘charity’.

Question. Check whether x3 – 3x2 + 5x = (x – 2)3 is quadratic equation or not.
Solution. Here given equation is

which is of the form ax2 + bx + c = 0
Hence, given equation is a quadratic equation.

Question. Find the values of k for which the given equation has real and equal roots: x2 – 2x (1 + 3k) + 7(3 + 2k) = 0
Solution. Given, x2 – 2x(1 + 3k) + 7(3 + 2k) = 0
Here, a = 1, b = -2(1 + 3k) and c = 7(3 + 2k)
The given equation will have equal roots, if
D = 0
⇒ b2 – 4ac = 0

Question. Show that x = – 3 is a solution of x2 + 6x + 9 = 0. 6
Solution.

Question. Solve the quadratic equation by factorization: x2 – x – a(a +1) = 0
Solution. According to the question, x2 – x – a(a + 1) = 0

Question. There is a square field whose side is 44 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at Rs.2.75 and Rs. 1.50 per square metre, respectively, is Rs.4904. Find the width of the gravel path.
Solution. Let width of the gravel path be x m. Then,
Each side of the square flower bed = (44 – 2x) m.
Area of the square field = 44 x 44 = 1936 m2

Area of the flower bed = (44 – 2x)(44 – 2x) = (44 – 2x)2 m2
∴ Area of the gravel path = Area of field – Area of flower bed
= 1936 – (44 – 2x)2
= 1936 – (1936 – 176x + 4x2)
= 1936 – 1936 + 176x – 4x2
= (176x – 4x2) m2
Cost of laying the flower bed = (Area of flower bed) (Rate per sq. m)
According to question,
∴ 11(22-x)2+6(44x-x2)=4904
⇒ 11(484 – 44x + x2) + (264x – 6x2) = 4904
⇒ 5x2 – 220x + 5324 = 4908
⇒ 5x2 – 220x + 420 = 0
⇒ x2 – 44x + 84 = 0
⇒ x2 – 42x – 2x + 84 = 0
⇒ x(x-42)-2(-42)=0
⇒ (x-2)(x-42)=0
⇒ x=2 or x=42
But, x ≠ 42, as the side of the square is 44 m. Therefore, x = 2.
Hence, the width of the gravel path is 2 metres.

Question. If x = – 2 is a root of the equation 3x2 + 7x + p = 0, find the values of k so that the roots of the equation x2 + k(4x + k – 1) + p = 0 are equal.
Solution. Here, x = – 2 is a root of 3x2 + 7x + p = 0
⇒ 3(-2)2 + 7 (-2) + p = 0
⇒ p = 2
∴ x2 + k(4x + k – 1) + p = 0 becomes
x2 + 4kx + k2 – k + 2 = 0….(i)
Comparing eq. (i) with ax2 + bx + c = 0 , we get
a = 1 , b = 4k and c = k2 – k + 2
Roots of eq (i) are equal
So, D =b2 – 4ac =0
⇒ (4k)2 – 4 x 1 (k2 – k + 2) = 0
⇒ 16k2 – 4k2 + 4k – 8 = 0
⇒ 12k2 + 4k – 8 = 0
⇒ 12k2 + 4k – 8 = 0
⇒ 3k2 + k – 2 = 0
⇒ 3k2 + 3k – 2k – 2 = 0
⇒ 3k(k + 1) – 2(k + 1) = 0
⇒ (k + 1)(3k – 2) = 0
⇒ k = -1, k = 2/3

Question. If – 2 is a root of the equation 3x2 – 5x + 2k = 0, find the value of k.
Solution. We have the following equation,

Question. A man buys a number of pens for Rs. 180. If he had bought 3 more pens for the same amount, each pen would have cost him Rs. 3 less. How many pens did he buy?
Solution. Let the number of pens purchased be x.
Cost of 1 pen = Rs. 180/x
If number of pens increase by 3. Then,
Cost of one pen = Rs. 180 / x +3
According to question,

⇒ 540 = 3x2 + 9x
⇒ 3x2 + 9x – 540 = 0
⇒ x2 + 3x – 180 = 0
⇒ x2 + 15x – 12x – 180 = 0
⇒ x(x + 15) – 12(x + 15) = 0
⇒ x + 15 = 0 or x – 12 = 0
⇒ x = -15 or x = 12
As number of pens can’t be negative.
⇒ x = 12
Therefore, he bought 12 pens.

Question. Find the discriminant of equation: 2x2 – 7x + 6 = 0.
Solution. Given, 2x2 – 7x + 6 = 0
∴ a =2, b = -7 and c = 6
D = b2 -4ac
= (-7)2 – 4(2)(6)
= 49 – 48
= 1

Question. Solve the following problem: x2 − 45x + 324 = 0
Solution. x2 – 45x + 324 = 0
⇒ x2 – 36x – 9x + 324 = 0 ⇒ x (x – 36) – 9(x – 36)=0
⇒ (x – 9)(x – 36) ⇒ x = 9, 36

Question. Find the roots of the equation, if they exist, by applying the quadratic formula: x2 + 5x – (a2 + a – 6) = 0.
Solution.

Question. At t minutes past 2 p.m, the time needed by the minute hand of a clock to show 3 p.m. was found to be 3 minutes less than t2/4 minutes. Find t.
Solution. Total time taken by minute hand from 2 p.m. to 3 p.m. is 60 min.
According to question,
t + (t2/4 – 3) = 60
⇒ 4t + t2 – 12 = 240
⇒ t2 + 4t – 252 = 0
⇒ t2 + 18t – 14t – 252 = 0
⇒ t(t + 18) – 14(t +18) = 0
⇒ (t + 18) (t – 14) = 0
⇒ t + 18 = 0 or t – 14 = 0
⇒ t = –18 or t = 14 min.
As time can’t be negative. Therefore, t = 14 min.

Question. Use factorization method to solve the quadratic equation ad2x(a/bx + 2c/d) + c2b = 0
Solution.

Question. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.
Solution. Let the ten’s place digit be y and unit’s place be x.
Therefore, number is 10y + x.
According to given condition,
10y + x = 4(x + y) and 10y + x = 2xy
⇒ x = 2y and 10y + x = 2xy
Putting x = 2y in 10y + x = 2xy
10y + 2y = 2.2y.y
12y = 4y2
4y2 – 12y = 0 ⇒ 4y(y – 3) = 0
⇒ y – 3 = 0 or y = 3
Hence, the ten’s place digit is 3 and units digit is 6 (2y = x)
Hence the required number is 36.

Question. Sum of the areas of two squares is 400 cm . If the difference of their perimeters is 16 cm, find the sides of the two squares.
Solution. Let the sides of two squares be a and b,
then a2 + b2 = 400 …(i)
and 4(a -b ) = 16
or, a -b = 4:
or, a =4 + b (ii)
From equations (i) and (ii), we get

Question. Without solving, find the nature of the roots of the quadratic equations. x2 + x + 1 = 0.
Solution.

Question. Solve: 1/(x + 3) + 1/(2x – 1) = 11/(7x + 9) , x ≠ -3 , 1/2 , -9/7
Solution. Given,

Now cross multiply
⇒ (3x + 2)(7x + 9) = 11(2x2 + 5x – 3)
⇒ 21x2 + 41x +18 = 22x2 + 55x – 33
⇒ x2 + 14x – 51 = 0
⇒ x2 + 17x – 3x – 51 = 0
⇒ x(x + 17) – 3(x + 17) = 0
⇒ (x + 17)(x – 3) = 0
⇒ x + 17 = 0 or x – 3 = 0
⇒ x = -17 or x = 3.
Therefore, -17 and 3 are the roots of the given equation.

Question. Solve for x: x – 1 /x – 2 + x – 3 / x – 4 = 3 , 1/3 (x ≠ 2 , 4)
Solution. The given equation is

Hence, the solutions of the given equation and 5 and 5/2

Question. Solve: x2 + 6x + 5 = 0
Solution. Given, x2+ 6x + 5 = 0
Splitting middle term,

Question. Check whether it is quadratic equation: (x + 1)3 = x3 + x + 6
Solution. We have the following equation,
(x + 1)3 = x3 + x + 6
⇒ x3 + 1 + 3x(x + 1) = x3 + x + 6
⇒ 3x2 + 2x – 5 = 0.
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.

Question. The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle?
Solution. Suppose, the smaller side of the right triangle be x cm and the larger side be y cm.
Then,

If the smaller side is tripled and the larger side be doubled, the new hypotenuse is 9√5 cm.

But, length of a side can not be negative. Therefore, y = 9
Hence, the length of the smaller side is 3 cm and the length of the larger side is 9 cm.

Question. Find the roots of the quadratic equation 2x2 – x – 6 = 0
Solution. Given, 2x2 – x – 6 = 0
Splitting the middle term of the equation,
⇒ 2x2 – 4x + 3x – 6 = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (x – 2)(2x + 3) = 0
⇒ x – 2 = 0 or 2x + 3 = 0
Therefore, x = 2 or x = -3/2

Question. A journey of 192 km from a town A to town B takes 2 hours more by an ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speed of the faster and the passenger train.
Solution.

or x(x + 48) – 32(x + 48) = 0
or, (x – 32) (x + 48) = 0
or, x = 32 or – 48
Since speed can’t be negative, therefore – 48 is not possible.
∴ Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h

Question. Solve for x: (2x / x -5)2 + 5 (2x /x – 5 ) – 24 = 0 , x ≠ 5
Solution. We have given,

Question. A motor boat, whose speed is 15 km/hr in still water, goes 30 km downstream and comes back in a total time of 4 hr 30 minutes, find the speed of the stream.
Solution. Speed of motor boat in still water = 15 km/hr
Speed of stream = x km/hr
Speed in downward direction = 15 + x
Speed in upward direction = 15 – x
According to question,

Question. If 2x2 – (2 + k)x + k = 0 where k is a real number, find the roots of the equation.
Solution. Given quadratic equation is 2x2 – (2 + k)x + k = 0 .

Question. A train travels a distance of 480 km’s at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. Find the quadratic equation in terms of the speed of the train.
Solution. Distance travelled by the train = 480 km
Let the speed of the train be x kmph
Time taken for the journey = 480/x
Given speed is decreased by 8 kmph
Hence the new speed of train = (x – 8) kmph
Time taken for the journey = 480 / x – 8

On solving we get x = 40
Thus the speed of train is 40 kmph.

Question. Solve the following problem: x2−55x +750=0
Solution. x 2−55x +750=0
⇒ x 2−25x−30x +750=0 ⇒ x (x−25)- 30(x−25)=0
⇒ (x−30)(x−25) ⇒ x =30, 25

Question. Solve the quadratic equation x2 + (a/a+b + a + b ) x + 1 = 0 by factorization.
Solution.

Question. Check whether(x – 7)x = 3x2 – 5 is a quadratic equation:
Solution. Given equation is (x – 7) x = 3x2 – 5
⇒ x2 – 7x = 3×2 -5
⇒ 2x2 + 7x – 5 = 0
which of the form ax2 + bx + c = 0
Hence, given equation is a quadratic equation.

Question. A water tank is connected with three pipes of uniform flow. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. If the fist two pipes operating simultaneously, fill the tank in the same time as the time taken by the third pipe alone to fill the tank. Find the time taken by each pipe individually to fill the tank.
Solution. Let the time taken by the second pipe to fill the tank individually be x hrs.
Then, the time taken by the first and third pipes to fill the tank individually are (x +5)
hrs and (x – 4) hrs respectively.

Either x = 10 or x = – 2
But x cannot be negative. So, x = 10
Hence, the time taken by first, second and third pipe to fill the tank individually are
15 hours, 10 hours and 6 hours respectively.

Question. What is the nature of roots of the quadratic equation 5x2 – 2x – 3 = 0?
Solution. On comparing equation with standard form of equation i.e, ax2 + bx + c = 0, we get
a = 5, b = -2, c = -3
Now , D = b2 – 4ac = (-2)2 – 4 x 5 x (-3)
= 4 + 60 = 64
Therefore, D = 64
We know , For D>0 , the roots of equation are real and distinct.
Therefore , 5x2 – 2x – 3 = 0 has real and distinct roots.

Question. A farmer wishes to grow a 100 m2 rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side-fence. Find the dimensions of his garden.
Solution. Let the length of one side be x metres and other side be y metres.
Then, x + y + x = 30 y = 30 ⇒ – 2x
Area of the garden = 100 m2
⇒ xy = 100
⇒ x (30-2x) = 100
⇒ 2x2 – 30x +100 = 0
⇒ 2(x2-15x+50) = 0 or x2-15x+50 = 0
⇒ x2-10x – 5x +50 = 0
⇒ x (x-10) – 5 (x-10) = 0
⇒ (x-10) (x-5) = 0
Either x-10 = 0 or x-5 = 0
⇒ x= 10, 5
∴ y = 30 – 20 = 10 or 30 -10 =20
Hence, the dimensions of the vegetable garden are 5m 20m or 10m 10m.

Question. Solve for x : 2x / x – 3 + 1 / 2x+ 3  + 3x + 9/(x – 3)(2x + 3) = 0 , x ≠ 3 , – 3/2
Solution.

Question. Determine whether the given values are solution of the given equation or not: 6x2 – x – 2 = 0, x = -1/2, x = 2/3
Solution. We have,  6x2 – x – 2 = 0
Substituting x = -1/2 , we get

Question. If x = -1/2 , is a solution of the quadratic equation 3x2 + 2kx – 3 = 0, find the value of k.
Solution. We have,
3x2 +2kx – 3 = 0
since x = – 1/2

Question. Solve for x by quadratic formula p2x2 + (p2 – q2)x – q2 = 0   17
Solution. p2x2 + (p2 – q2)x – q2 = 0
a = p2, b = p2 – q2, c = -q2
D = b2 – 4ac
= (p2 – q2)2 – 4 x p2(-q2)
= p4 + q2 – 2p2q2 + 4p2q2
= (p2 + q2)2

Question. A faster train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.
Solution. 40 km/hr, 50 km/hr

Question. X and Y are centres of circle of radius 9 cm and 2 cm and XY = 17 cm. Z is the centre of a circle of radius r cm, which touches the above circles externally. Given that ∠XZY = 90°, find the value of r.
Solution. r = 6 cm

Question. A two digit number is such that the product of its digit is 10. When 27 is subtracted from the number, the digits interchange their places. Find the number.
Solution. 52

Question. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution. 36

Question. Find the values of k so that the equation (k + 2) x2 – (7 – k) x + 3 = 0 has :
(a) equal roots (b) distinct roots (c) no real roots.
Solution. (a) k = 1, 25 (b) k > 25 or k < 1 (c) 1 < k < 2

Question. The sum of the ages of a man and his son is 40 years. The product of their ages is 144 years. Find their present ages.
Solution. 36 years and 4 years

Question. The sum of two numbers is 16. The sum of their reciprocals is 1/3 Find the numbers.
Solution. 4, 12

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.
Solution. 1 hour

Question. The length of a rectangular ground is greater than twice its breadth by 5 m. The area of the ground is 1125 sq. m. Find the length and breadth of the ground.
Solution. length = 50 m, breadth = 45/2 m

Question. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the triangle be 60 sq. cm., calculate the length of the sides of the triangle.
Solution. 15 cm, 8 cm, 17 cm

Question. The difference of the squares of two numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Solution. 6 and 9

Question. A carpet is placed in a room 6m × 4m, leaving a border of a uniform width all around it. Find the width of the border, if the area of the carpet is 8 sq. m.
Solution. 1 m

Question. Two circles touch externally. The sum of their areas is 130 π sq.cm and the distance between their centres is 14 cm. Find the radii of the circles.
Solution. 11 cm and 3 cm

Question. The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 sq. m. Find its dimensions.
Solution. 25 m × 20 m

Question. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice th area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Find the sides of the squares.
Solution. 5 cm and 8 cm

Question. In an auditorium, seats are arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row is reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Solution. (i) 30 (ii) 1200

Question. The area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm. Find its base.
Solution. 10 cm or 24 cm

Question. The sum of the squares of two consecutive odd positive integers is 290. Find them.
Solution. 11, 13

Question. A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. How many books did he buy?
Solution. 16

Question. A line AB is 8 cm in length. AB is produced to P such that BP2 = AB. AP. Find the length of BP.
Solution. 4 (1 + √5) cm

Question. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy.
Solution. Rs. 20

Question. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution. Swati’s present age = 9 years, varun’s present age = 27 years

Question. A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.
Solution. 20 days

Question. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hour 30 minutes. Find the speed of the stream.
Solution. 5 km/hr

Question. Rs. 6500 were divided among a certain number of persons. Had there been 15 more persons, each would have got Rs. 30 less. Find the original number of the persons.
Solution. 50

Question. A piece of cloth costs Rs. 200. If the piece was 5 m longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Solution. 20 m, Rs. 10

Question. The denominator of a fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is 3 , 11/14 .Find the fraction.
Solution. 2/7

Question. A takes 15 days less than the time taken by B to finish piece of work. If both A and B together can finish the work in 18 days, find the time taken by B to finish the work.
Solution. 45 days

Question. If the sum of first n even natural numbers is 420, find the value of n.
Solution. n = 20

Question. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Solution. 25 km/hr

Question. Two pipes running together can fill a cistern in 3 , 1/3 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Solution. 5 minutes and 8 minutes

Question. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find the speed of the stream.
Solution. 3 km/hr

Question. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution. 10, 6

Question. In a class test, the sum of Nishu’s marks in Mathematics and Science is 30. Had she got 2 marks more in Mathematics and 3 marks less in Science, the product of their marks would have been 210. Find her marks in the two subjects.
Solution. Marks in mathematics = 13, Marks in science = 17 OR Marks in Mathematics = 12, Marks in science = 18

Question. Find two consecutive numbers whose squares have the sum 85.
Solution. 6, 7

Question. A polygon of n side has n(n – 3) diagonals. How many sides has a polygon with 54 diagonals?
Solution. 12

Question. Two numbers differ by 3 and their product is 504. Find the numbers.
Solution. 21, 24 or – 24, – 21

Question. A farmer wishes to start a 300 sq. m rectangular vegetable garden. Since he has only 50 m barbed wire, he fences three sides of the rectangular garden, letting his house compound wall act as the fourth side fence. Find the dimensions of the garden.
Solution. 20 m × 15 m or 30 m × 10 m

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 sq. m more than the area of a park that was already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find its length and breadth.

Solution. 7 m, 4 m

Question. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his sons age. Find their present ages.
Solution. 7 years and 49 years

Question. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Solution. 150 cm2

Question. Sum of the areas of two squares is 468 m2. If the difference of their perimeter is 24 m, find the sides of the two squares.
Solution. 18 m and 12 m

Question. O girl! Out of a group of swans, 7/2 times the square root of the number are playing on the shore of a tank. The two remaining ones are playing with amorous fight, in the water. What is the total number of swans?
Solution. 16

Question. The angry Arjun carried some arrows for fighting with Bheesm. With half the arrows, he cut down the arrows thrown by Bheesm on him and with six other arrows, he killed the rath driver of the Bheesm. With one arrow each, he knocked down respectively the rath, flag and the bow of Bheesm. Finally, with one more than four times the square root of the total number of arrows, he laid Bheesm unconscious on an arrow bed. Find the total number of arrows Arjun had.
Solution. 100

Question. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If, after 2 hours, they are 50 km apart, find the speed of each trains.
Solution. 20 km/hr, 15 km/hr

Question. In an auditorium, the number of rows was equal to number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. How many rows were there?
Solution. 30 rows

Question. Some students planned a picnic. The budget the food was Rs. 500. But, 5 of them failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic?
Solution. 20

Question. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km per hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Solution. 45 km/hr

Question. Out of a number of Saras birds, one fourth the number are moving about in lotus plants; 1/9 th coupled (along) with 1/4 th as well as 7 times the square root of the number move on a hill; 56 birds remain in vakula trees. What is the total number of birds?
Solution. 576

Question. A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digits are reversed. Find the number.
Solution. 27

Question. The length of the hypotenuse of a right-angled triangle exceeds the base by 1 cm and also exceeds twice the length of the altitude by 3 cm. Find the length of each side of the triangle.
Solution. 5 cm, 12 cm, 13 cm

Question. Find the values of k for which the equation x2 + 5kx + 16 = 0 has no real roots.
Solution. – 8 /5 <k< 8/5

Question. Solve the equation 2x2 – 5x + 3 = 0 by the method of completing the square.
Solution. 1 , 3/2

Question. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Solution. 4, 12

Question. A two digit number is four times the sum of its digits and twice the product of its digits. Find the number.
Solution. 36

Question. Solve for x using quadratic formula :
a2b2x2 + b2x – a2x – 1 = 0
Solution. – 1/ a2 , 1/b2

Question. Divide 26 into two parts, whose sum of squares is 346. Frame an equation for the given statement.
Solution. x2 + (26 – x)2 = 346

Question. Solve for x : 1/ x + 4 – 1/x – 7 = 11/30 ; x ≠ – 4, 7
Solution. 1, 2

Question. Solve for x : 4x2 – 4ax + a2 – b2 = 0.
Solution. a + b /2 , x = a – b /2

Question. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Solution. 800 km/hr

Question. Find the values of k so that the quadratic equation x2 – 2x (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.
Solution. 2 or – 10/9

Question. Represent the following situation in the form a quadratic equation:
(a) Abdul and Sneha together have 30 oranges. Both of them ate 3 oranges each and the product of the number of oranges they have now is 120. We would like to find out how many oranges they had initially.
(b) The area of a rectangular plot is 428 m2. The length of the plot (in metres) is two more than twice its breadth. We need to find the length and breadth of the plot.
Solution. (a) Let the number of number of left oranges with Abdul and Sneha be x and y respectively.
Then, x + y = 30 ⇒ y = 30 – x
The number of oranges left with both Abdul and Sneha are x – 3 and y – 3 respectively.
The product of number of left oranges = 120
⇒ (x – 3) (y – 3) = 120
⇒ (x – 3) (30 – x – 3) = 120 (Q y = 30 – x)
⇒ (x – 3) (27 – x) = 120
⇒ 27x – x2 – 81 + 3x = 120 ⇒ x2 – 30x + 201 = 0
(b) Let the breadth of the plot be x. Then the length of the plot = 2x + 2
Since, area of the plot = 428 m2 (Given)
∴ x(2x + 2) = 428
⇒ 2x2 + 2x – 428 = 0
⇒ x2 + x – 214 = 0

Question. If 1/2 is a root of the equation x2 + kx – 5/4 = 0 , then find the value of k.
Solution. It is given that 1/2 is a root of quadratic equation.
∴ It must satisfy the quadratic equation:

Question. If a and b are the roots of the equation x2 + ax – b = 0, then find a and b.
Solution. Here, A = 1, B = a, C = –b

Question. Check whether the following are quadratic equations:
(a) x(x + 2) – 3 = (x + 4) x (b) (x + 2)3 = x3 – 4x2 + 2
Solution. (a) Since x(x + 2) – 3 = x(x + 4)
⇒ x2 + 2x – 3 = x2 + 4x
⇒ 2x + 3 = 0
This is linear equation not a quadratic equation.
(b) (x + 2)3 = x3 – 4x2 + 2
⇒ x3 + 6x2 + 12x + 8 = x3 – 4x2 + 2
⇒ 10x2 + 12x + 6 = 0
⇒ 5x2 + 6x + 3 = 0
This is a quadratic equation.

Question. Determine whether 3 is a root of the equation

Solution.

Question. Write the set of values of k for which the quadratic equation 2x2 + kx + 8 has real roots.
Solution. For real roots, D ≥ 0
⇒ b2 – 4ac ≥ 0 ⇒ k2 – 4(2) (8) ≥ 0
⇒ k2 – 64 ≥ 0 ⇒ k2 ≥ 64 ⇒ k < –8 and k > 8

Question. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.
Solution. Let the roots of the given equation be a and 6a.
Thus the quadratic equation is (x – a)(x – 6a) = 0
⇒ x2 – 7ax + 6a2 = 0 …(i)
Given equation can be written as
x2 – 14/P x + 8/P = 0
Comparing the coefficients in (i) and (ii) 7α= 14/P and 6α2 = 8/P
Solving to get p = 3.

Question. State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution. We have (x + 1) (x – 2) + x = 0 ⇒ x2 – x – 2 + x = 0 ⇒ x2 – 2 = 0
∴ D = b2 – 4ac = 0 – 4(1) (–2) = 8 > 0
∴ Given equation has two distinct real roots.

Question. If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.
Solution. ∴ –5 is a root of the equation 2×2 + px – 15 = 0
∴ 2 (–5)2 + p (–5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p (x2 + x) + k = 0 or 7x2 + 7x + k = 0 has equal roots
∴ D = 0
i.e., b2 – 4ac = 0 or 49 – 4 × 7k = 0 ⇒ k = 49/28 = 7/4

Question. If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots, show that c2 = a2 (1 + m2).
Solution. The given equation is (1 + m2) x2 + (2mc) x + (c2 – a2) = 0
Here, A = 1 + m2, B = 2mc and C = c2 – a2
Since the given equation has equal roots, therefore
D = 0 ⇒ B2 – 4AC = 0
⇒ (2mc)2 – 4 (1 + m2) (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0
⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0 [Dividing throughout by 4]
⇒ –c2 + a2 (1 + m2) = 0 ⇒ c2 = a2 (1 + m2) Hence Proved.

Question. Find the value of k for the quadratic equation kx (x – 2) + 6 = 0, so that it has two equal roots.
Solution. We have kx (x – 2) + 6 = 0
⇒ kx2 – 2kx + 6 = 0
Here, a = k, b = –2k, c = 6
For equal roots, D = 0
i.e., b2 – 4ac = 0 ⇒ (–2k)2 – 4 × k × 6 = 0
⇒ 4k2 – 24k = 0 ⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0 ⇒ k = 0 or k = 6
But k ≠ 0 (because if k = 0, then given equation will not be a quadratic equation).
So, k = 6.

Question. Form a quadratic equation whose roots are 2 and 3.
Solution. Sum of roots = 2 + 3 = 5
Product of roots = 2 × 3 = 6
So, quadratic equation can be formed as x2 – (sum of roots)x + product of roots = 0
x2 – 5x + 6 = 0

Question. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(a) 3x2 – 4√3x + 4 = 0 (b) 2x2 – 6x + 3 = 0
Solution. (a) We have, 3×2 − 4√3x + 4 = 0
Here, a = 3, b = −4 3 and c = 4
Therefore, D = b2 – 4ac = (−4 √3) − 4 × 3 × 4 = 48 − 48 = 0
Hence, the given quadratic equation has real and equal roots.

(b) We have, 2x2 – 6x + 3 = 0
Here a = 2, b = –6, c = 3
Therefore, D = b2 – 4ac = (–6)2 – 4 × 2 × 3 = 36 – 24 = 12 > 0
Hence, given quadratic equation has real and distinct roots.

Question. What will be the nature of roots of quadratic equation 2x2 + 4x – 7 = 0?
Solution. ∴ 2x2 + 4x – 7 = 0
Here, a = 2, b = 4, c = –7
D = b2 – 4ac = 16 – 4 × 2 × (–7) = 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Question. Find the value(s) of k for which the equation x2 + 5kx + 16 = 0 has real and equal roots.
Solution. For roots to be real and equal, b2 – 4ac = 0
⇒ (5k)2 – 4 × 1 × 16 = 0 ⇒ k = ± 8/5

Question. If the roots of the equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.
Solution. Since the equation (a – b) x2 + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)2 – 4 (a – b) (c – a) = 0
⇒ b2 + c2 – 2bc – 4 (ac – a2 – bc + ab) = 0
⇒ b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
⇒ 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0
⇒ (2a)2 + (–b)2 + (–c)2 + 2 (2a) (–b) + 2 (–b) (–c) + 2 (–c) 2a = 0
⇒ (2a – b – c)2 = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved.

Question. If ax2 + bx + c = 0 has equal roots, find the value of c.
Solution. For equal roots D = 0
i.e., b2 – 4ac = 0 ⇒ b2 = 4ac ⇒ c = b2/4a

## Mathematics Quadratic Equation Worksheets for Class 10 as per CBSE NCERT pattern

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