## Worksheet for Class 10 Mathematics Chapter 4 Quadratic Equation

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 4 Quadratic Equation in Grade 10. This test paper with questions and solutions for Standard 10 Mathematics will be very useful for tests and exams and help you to score better marks

### Chapter 4 Quadratic Equation Class 10 Mathematics Worksheet Pdf

**VERY SHORT ANSWER TYPE QUESTIONS :**

**Question. If one of the roots of x ^{2} + px − 4 = 0 is − 4 then find the product of its roots and the value of p.**

**Answer :**If − 4 is a root of the quadratic equation,

x

^{2}+ px − 4 = 0

∴ (− 4)

^{2}+ (− 4) (p) − 4 = 0

⇒ 16 − 4p − 4 = 0

⇒ 12 − 4p = 0 ⇒ p = 3

Now, in ax

^{2}+ bx + c = 0, the product of the roots = c/a

∴ Product of the roots in x

^{2}− px − 4 = 0

= − 4/1 = − 4

**Question. For what value of k are the roots of the quadratic equation kx ^{2} + 4x + 1 = 0 equal and real?**

**Answer :**Comparing kx

^{2}+ 4x + 1 = 0, with ax

^{2}+ bx + c = 0, we get

a = k

b = 4

c = 1

∴ b

^{2}− 4ac = (4)

^{2}− 4 (k) (1)

= 16 − 4k

For equal and real roots, we have

b

^{2}− 4ac = 0

⇒ 16 − 4k = 0

⇒ 4k = 16

⇒ k = 16/4 = 4

**Question. Using quadratic formula, solve the following quadratic equation for x:****x ^{2} − 2ax + (a^{2} − b^{2}) = 0**

**Answer :**Comparing x

^{2}− 2ax + (a

^{2}− b

^{2}) = 0, with ax

^{2}+ bx + c = 0, we have:

a = 1, b = − 2a, c = a

^{2}− b

^{2}

∴ x = − b ± √b

^{2}− 4ac/2a

⇒ x = − (− 2a) ± − √(2a)

^{2}− 4 (1) (a

^{2}− b

^{2})/2)1_

= 2a ± √4a

^{2}− 4a

^{2 }− 4b

^{2}

= 2a ± √4b

^{2}/2 = 2a ± b/2 = a ± b

∴ x = (a + b) or x = (a − b)

**Question. For what value of k, does the quadratic equation x ^{2} − kx + 4 = 0 have equal roots?**

**Answer :**Comparing x

^{2}− kx + 4 = 0 with ax

^{2}+ bx + c = 0, we get

a = 1

b = − k

c = 4

∴ b

^{2}− 4ac = (− k)

^{2}− 4 (1) (4) = k

^{2}− 16

For equal roots,

b

^{2}− 4ac = 0

⇒ k

^{2}− 16 = 0

⇒ k

^{2}= 16

⇒ k = ± 4

**Question. If the roots of the quadratic equation − ax ^{2} + bx + c = 0 are equal then show that b^{2} = 4ac.**

**Answer :**ä For equal roots, we have

b

^{2}− 4ac = 0

∴ b

^{2}= 4ac

**Question. If one root of the quadratic equation 2x ^{2} − 3x + p = 0 is 3, find the other root of the quadratic equation. Also find the value of p.**

**Answer :**We have:

2x

^{2}− 3x + p = 0

∴ a = 2, b = − 3 and c = p

Since, the sum of the roots = − b/a

= − (−3)/2 = 3/2

∴ One of the roots = 3

∴ The other root = 3/2 − 3 = − 3/2

Now, substituting x = 3 in (1), we get

2 (3)

^{2}− 3 (3) + p = 0

⇒ 18 − 9 + p = 0

⇒ 9 + p = 0 ⇒ p = − 9

**Question. For what value of k does (k − 12) x ^{2} + 2 (k − 12) x + 2 = 0 have equal roots?**

**Answer :**Comparing (k − 12) x

^{2}+ 2 (k − 12) x + 2 = 0 with ax

^{2}+ bx + c = 0, we have:

a = (k − 12)

b = 2 (k − 12)

c = 2

∴ b

^{2}− 4ac = [2 (k − 12)]

^{2}− 4 (k − 12) (2)

= 4 (k − 12)

^{2}− 8 (k − 12)

= 4 (k − 12) [k − 12 − 2]

= 4 (k − 12) (k − 14)

For equal roots,

b

^{2}− 4ac = 0

⇒ 4 (k − 12) [k − 14] = 0

⇒ Either 4 (k − 12) = 0 ⇒ k = 12

or k − 14 = 0 ⇒ k = 14

But k = 12 makes k − 12 = 0 which is not required

∴ k ≠ 12

⇒ k = 14

**Question. For what value of k, does the given equation have real and equal roots?****(k + 1) x ^{2} − 2 (k − 1) x + 1 = 0.**

**Answer :**Comparing the given equation with ax

^{2}+ bx + c = 0, we have:

a = k + 1

b = − 2 (k − 1)

c = 1

For equal roots, b

^{2}− 4ac = 0

∴ [− 2 (k − 1)]

^{2}− 4 (k + 1) (1) = 0

⇒ 4 (k − 1)

^{2}− 4 (k + 1) = 0

⇒ 4 (k

^{2}+ 1 − 2k) − 4k − 4 = 0

⇒ 4k

^{2}+ 4 − 8k − 4k − 4 = 0

⇒ 4k

^{2}− 12k = 0

⇒ 4k (k − 3) = 0

⇒ k = 0 or k = 3

**Question. For what value of k does the equation 9x ^{2} + 3kx + 4 = 0 has equal roots?**

**Answer :**Comparing 9x

^{2}+ 3kx + 4 = 0 with ax

^{2}+ bx + c = 0, we get

a = 9

b = 3k

c = 4

∴ b

^{2}− 4ac = (3k)

^{2}− 4 (9) (4)

= 9k

^{2}− 144

For equal roots,

b

^{2}− 4ac = 0

⇒ 9k

^{2}− 144 = 0

⇒ 9k

^{2}= 144

⇒ k

^{2}= 144/9 = 16

⇒ k = ± 4

**Question. What is the nature of roots of the quadratic equation 4x ^{2} − 12x + 9 = 0?**

**Answer :**Comparing 4x

^{2}− 12x + 9 = 0 with ax

^{2}+ bx + c = 0 we get

a = 4

b = − 12

c = 9

∴ b

^{2}− 4ac = (− 12)

^{2}− 4 (4) (9)

= 144 − 144 = 0

Since b

^{2}− 4ac = 0

∴ The roots are real and equal.

**Question. If one of the roots of the quadratic equation 2x ^{2} + kx − 6 = 0 is 2, find the value of k. Also find the other root.**

**Answer :**Given equation:

2x

^{2}+ kx − 6 = 0

one root = 2

Substituting x = 2 in 2x

^{2}+ kx − 6 = 0

We have:

2 (2)

^{2}+ k (2) − 6 = 0

⇒ 8 + 2k − 6 = 0

⇒ 2k + 2 = 0 ⇒ k = − 1

∴ 2x

^{2}+ kx − 6 = 0 ⇒ 2x

^{2}− x − 6 = 0

Sum of the roots = − b/a = 1/2

∴ other root = 1/2 − 2

= 3/2

**Question. Find the value of ‘k’ for which the quadratic equation kx ^{2} − 5x + k = 0 have real roots.**

**Answer :**Comparing kx

^{2}− 5x + k = 0 with ax

^{2}+ bx + c = 0, we have:

a = k

b = − 5

c = k

∴ b

^{2}− 4ac = (− 5)

^{2}− 4 (k) (k)

= 25 − 4k

^{2}

For equal roots, b

^{2}− 4ac = 0

∴ 25 − 4k

^{2}= 0

⇒ 4k

^{2}= 25

⇒ k

^{2}= 25/4

⇒ k = ±√25/4 = ±5/2

**Question. Write the value of k for which the quadratic equation x ^{2} − kx + 9 = 0 has equal roots.**

**Answer :**Comparing x

^{2}− kx + 9 = 0 with ax

^{2}+ bx + c = 0, we get

a = 1

b = − k

c = 9

∴ b

^{2}− 4ac = (− k)

^{2}− 4 (1) (9)

= k

^{2}− 36

For equal roots,

b

^{2}− 4ac = 0

⇒ k

^{2}− 36 = 0 ⇒ k

^{2}= 36

⇒ k = ± 6

**Question. If − 4 is a root of the quadratic equation x ^{2} + px − 4 = 0 and x^{2} + px + k = 0 has equal roots, find the value of k.**

**Answer :**∴ (–4) is a root of x

^{2}+ px − 4 = 0

∴ (− 4)

^{2}+ p (− 4) = 0

⇒ 16 − 4p − 4 = 0

⇒ 4p = 12 or p = 3

Now, x

^{2}+ px + k = 0

⇒ x

^{2}+ 3x + k = 0 [ ∴ p = 3]

Now, a = 1, b = 3 and c = + k

∴ b

^{2}− 4ac = (3)2 − 4 (1) (k)

= 9 − 4k

For equal roots, b

^{2}− 4ac = 0

⇒ 9 − 4k = 0 ⇒ 4k = 9

⇒ k = 9/4

**Question. Determine the value of k for which the quadratic equation 4x ^{2} − 4kx + 1 = 0 has equal roots.**

**Answer :**We have:

4x

^{2}− 4kx + 1 = 0

Comparing with ax

^{2}+ bx + c = 0, we have

a = 4, b = − 4k and c = 1

∴ b

^{2}− 4ac = (− 4k)

^{2}− 4 (4k) (1)

= 16k

^{2}− 16

For equal roots

b

^{2}− 4ac = 0

∴ 16k

^{2}− 16 = 0

⇒ 16k

^{2}= 16 ⇒ k

^{2}= 1

⇒ k = ± 1

**Question. For what value of k are the roots of the quadratic equation 3x ^{2} + 2kx + 27 = 0 real and equal?**

**Answer :**Comparing 3x

^{2}+ 2 kx + 27 = 0 with ax

^{2}+ bx + c = 0, we have:

a = 3

b = 2k

c = 27

∴ b

^{2}− 4ac = (2k)

^{2}− 4 (3) (27)

= 4k

^{2}− (12 × 27)

For the roots to be real and equal

b

^{2}− 4ac = 0

⇒ 4k

^{2}− (12 × 17) = 0

⇒ 4k

^{2 }= 12 × 27

⇒ k

^{2}= 12 × 27/4 = 81

⇒ k = ± 9

**SHORT ANSWER TYPE QUESTIONS :**

**Question. Solve for x: 36x ^{2} − 12ax + (a^{2} − b^{2}) = 0.**

**Answer**: We have:

36x

^{2}− 12ax + (a

^{2}− b

^{2}) = 0

Comparing (1) with Ax

^{2}+ Bx + C = 0, we have:

A = 36

B = − 12a

C = (a2 − b2)

∴ B

^{2}− 4AC = [− 12a]

^{2}− 4 (36) [a

^{2}− b

^{2}]

= 144 a

^{2}− 144 (a

^{2}− b

^{2})

= 144 a

^{2}− 144 a

^{2}+ 144 b

^{2}

= 144 b

^{2}

Since, x = − B ± √B

^{2}− 4AC/2A

∴ x = − (− 12a) ± √144 b

^{2}/2 (36)

⇒ x = 12a ± 12b/72

Taking +ve sign, we have:

x = 12a + 12b/72

⇒ x = 12/72 (a + b) = 1/6 (a + b)

Taking −ve sign, we get

x = 12a − 12b/72 (a − b) = 1/6 (a − b)

Thus, the required roots are:

x = 1/6 (a + b) and x = 1/6 (a − b)

**Question. Find the roots of the quadratic equation 2x ^{2} - **√

**5x - 2 = 0, using the quadratic formula.**

**Answer :**

**Question. Find the roots of the equation:****1/x − 2 + 1/x = 8/2x + 5 : x ≠ 0, 2 −5/2****Answer :** We have:

**Question. Solve 2x ^{2} − 5x + 3 = 0.**

**Answer :**We have:

2x

^{2}− 5x + 3 = 0

Comparing (1) with ax2 + bx + c = 0,

∴ a = 2

b = − 5

c = 3

∴ b

^{2}− 4ac = (− 5)

^{2}− 4 (2) (3)

= 25 − 24 = 1

Since, x = − b ± √b

^{2}− 4ac/2a

∴ x = − (− 5) ± √1/2 (2)

= 5 ± 1/4

Taking, + ve sign,

x = 5 + 1/4 = 6/4 = 3/2

Taking, −ve sign,

x = 5 − 1/4 = 4/4 = 1

Thus, the required roots are

x = 3/2 and x = 1

**Question. Solve for x: 9x ^{2} − 6ax + a^{2} − b^{2} = 0.**

**Answer :**We have:

9x

^{2}− 6ax + (a

^{2}− b

^{2}) = 0

Comparing (1) with ax

^{2}+ bx + c = 0, we get

a = 9, b = − 6a and c = (a2 − b2)

∴ b

^{2}− 4ac = (− 6a)

^{2}− 4 (9) (a

^{2}− b

^{2})

= 36a

^{2}− 36 (a

^{2}− b

^{2})

= 36a

^{2}− 36a

^{2}+ 36b

^{2}

= 36b

^{2}= (6b)

^{2}

Since, x = − b ± √b

^{2}− 4ac/2a

∴ x = − (− 6a) ± 6√(6b)

^{2}/2(9)

⇒ x = 6a ± 6b/18

⇒ x = 6[a ± b]/18 = a ± b/3

a ± b a b

Taking the +ve sign, we get

x = a + b/3

Taking the −ve sign, we get

x = a − b/3

∴ The required roots are:

x = a + b/3 and x = a − b/3

**Question. Evaluate √20 + √20 + √20 + ****Answer :**

**Question. Find the roots of the equation:****1/x − 1/x − 3 = 4/3 ; x ≠ 0, 3****Answer :** We have:

**Question. Solve the following quadratic equation:****2x ^{2} + 4x − 8 = 0**

**Answer :**We have:

2x

^{2}+ 4x − 8 = 0

Dividing by 2, we get

x

^{2}+ 2x − 4 = 0

Comparing (1) with ax2 + bx + c = 0,

a = 1,

b = 2

c = − 4

∴ b

^{2}− 4ac = (2)

^{2}− 4 (1) (− 4)

= 4 + 16 = 20

Since, x = − b ± √b

^{2}− 4ac/2a

∴ x =

− 2 ± √20/2 (1)

⇒ x = − 2 ± 2√5/2

⇒ x = 2 [−1 ± √5]/2 = − 1 ± √5

Taking + ve sign, we get

x = [− 1 + √5]

Taking − ve sign, we get

x = [− 1 − √5]

Thus, the required roots are x = [− 1 + √5] and x = [− 1 − √5]

**Question. Using quadratic formula, solve the following quadratic equation for x:****x ^{2} − 4ax + 4a^{2} − b^{2} = 0**

**Answer :**Comparing the given equation with ax

^{2}+ bx + c = 0, we have:

a = 1

b = − 4a

c = 4a

^{2}− b

^{2}

∴ b

^{2}− 4ac = [− (4a)]

^{2}− 4 (1) [4a

^{2}− b

^{2}]

= 16a

^{2}− 4 (4a

^{2}− b

^{2})

= 16a

^{2}− 16a

^{2}+ 4b

^{2}

= 4b

^{2}= (2b)

^{2}

Since, x = − b ± √b

^{2}− 4ac/2a

∴ x = − (− 4a) ± √(2b)

^{2}/2 (1)

⇒ x = 4a ± 2b/2

⇒ x = 2/2 [2a ± b] = 2a ± b

Taking the +ve sign, x = 2a + b

Taking the −ve sign, x = 2a − b

Thus, the required roots are:

x = 2a + b and x = 2a − b

**Question. Using quadratic formula, solve the following quadratic equation for x:****x ^{2} − 2ax + (a^{2} − b^{2}) = 0**

**Answer :**Comparing the given equation with ax

^{2}+ bx + c = 0, we have:

**Question. Solve: 16x ^{2} − 8a^{2} x + (a^{4} − b^{4}) = 0 for x.**

**Answer :**We have:

16x

^{2}− 8a

^{2}x + a

^{4}− b

^{4}= 0

Comparing (1) with ax

^{2}+ bx + c = 0, we get

a = 16

b = − 8a

^{2}

c = (a

^{4}− b

^{4})

∴ b

^{2}− 4ac = [− 8a

^{2}]

^{2}− 4 (16) (a

^{4}− b

^{4})

= 64 a

^{4}− 64 (a

^{4}− b

^{4})

= 64 a

^{4}− 64 a

^{4}+ 64 b

^{4}

= 64 b

^{4}

**Question. Solve (using quadratic formula):****x ^{2} + 5x + 5 = 0**

**Answer :**We have:

x

^{2}+ 5x + 5 = 0

Comparing (1) with ax

^{2}+ bx + c = 0, we have:

a = 1

b = 5

c = 5

∴ b

^{2}− 4ac = (5)

^{2}− 4 (1) (5)

= 25 − 20 = 5

Since, x = − b ± √b

^{2}− 4ac/2a

∴ x = − 5 ± √5/2 (1)

⇒ x = − 5 ± √5/2

Taking +ve sign, we have:

x = − 5 + √5/2

Taking −ve sign, we have:

x = − 5 − √5/2

Thus, the required roots are:

x = − 5 + √5/2 and x = − 5 − √5/2

**Question. Find the roots of the following equation:****1/x + 4 − 1/x − 7 = 11/30; x = 4, 7****Answer :** We have:

1/x + 4 − 1/x − 7 = 11/30

**Extra Based Questions :**

**Check whether the following are quadratic equations:**

**Question. (x + 1) ^{2} = 2(x − 3)**

**Answer :**(x + 1)

^{2}= 2(x − 3)

We have:

(x + 1)2 = 2 (x − 3)

⇒ x

^{2}+ 2x + 1 = 2x − 6

⇒ x

^{2}+ 2x + 1 − 2x + 6 = 0

⇒ x

^{2}+ 7 = 0

Since x

^{2}+ 7 is a quadratic polynomial

∴ (x + 1)

^{2}= 2(x − 3) is a quadratic equation.

**Question. x ^{2} − 2x = (−2) (3 − x)**

**Answer :**x

^{2}− 2x = (− 2) (3 − x)

We have:

x

^{2}− 2x = (− 2) (3 − x)

⇒ x

^{2}− 2x = − 6 + 2x

⇒ x

^{2}− 2x − 2x + 6 = 0

⇒ x

^{2}− 4x + 6 = 0

Since x

^{2}− 4x + 6 is a quadratic polynomial

∴ x

^{2}− 2x = (−2) (3 − x) is a quadratic equation.

**Question. (x − 2) (x + 1) = (x − 1) (x + 3)****Answer :** (x − 2) (x + 1) = (x − 1) (x + 3)

We have:

(x − 2) (x + 1) = (x − 1) (x + 3)

⇒ x^{2} − x − 2 = x^{2} + 2x − 3

⇒ x^{2} − x − 2 − x^{2} − 2x + 3 = 0

⇒ −3x + 1 = 0

Since −3x + 1 is a linear polynomial

∴ (x − 2) (x + 1) = (x − 1) (x + 3) is not quadratic equation.

**Question. (x − 3) (2x + 1) = x(x + 5)****Answer :** (x − 3) (2x + 1) = x(x + 5)

We have:

(x − 3) (2x + 1) = x (x + 5)

⇒ 2x^{2} + x − 6x − 3 = x^{2} + 5x

⇒ 2x^{2} − 5x − 3 − x^{2} − 5x = 0

⇒ x^{2} + 10x − 3 = 0

Since x^{2} + 10x − 3 is a quadratic polynomial

∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.

**Question. (2x − 1) (x − 3) = (x + 5) (x − 1)****Answer :** (2x − 1) (x − 3) = (x + 5) (x − 1)

We have:

(2x − 1) (x − 3) = (x + 5) (x − 1)

⇒ 2x^{2} − 6x − x + 3 = x^{2} − x + 5x − 5

⇒ 2x^{2} − x^{2} − 6x − x + x − 5x + 3 + 5 = 0

⇒ x^{2} − 11x + 8 = 0

Since x^{2} − 11x + 8 is a quadratic polynomial

∴ (2x − 1) (x − 3) = (x + 5) (x − 1) is a quadratic equation.

**Question. x ^{2} + 3x + 1 = (x − 2)^{2}**

**Answer :**x

^{2}+ 3x + 1 = (x − 2)

^{2}

We have:

x

^{2}+ 3x + 1 = (x − 2)

^{2}

⇒ x

^{2}+ 3x + 1 = x

^{2}− 4x + 4

⇒ x

^{2}+ 3x + 1 − x

^{2}+ 4x − 4 = 0

⇒ 7x − 3 = 0

Since 7x − 3 is a linear polynomial.

∴ x

^{2}+ 3x + 1 = (x − 2)

^{2}is not a quadratic equation.

**Question. (x + 2) ^{3} = 2x(x^{2} − 1)**

**Answer :**(x + 2)

^{3}= 2x(x

^{2}− 1)

We have:

(x + 2)

^{3}= 2x(x

^{2}− 1)

⇒ x

^{3}+ 3x

^{2}(2) + 3x(2)

^{2}+ (2)

^{3}= 2x

^{3}− 2x

⇒ x

^{3}+ 6x

^{2}+ 12x + 8 = 2x

^{3}− 2x

⇒ x

^{3}+ 6x

^{2}+ 12x + 8 − 2x

^{3}+ 2x = 0

⇒ − x

^{3}+ 6x

^{2}+ 14x + 8 = 0

Since −x

^{3}+ 6x

^{2}+ 14x + 8 is a polynomial of degree 3

∴ (x + 2)

^{3}= 2x(x

^{2}− 1) is not a quadratic equation.

**Question. x ^{3} − 4x^{2} − x + 1 = (x − 2)^{3}**

**Answer :**x

^{3}− 4x

^{2}− x + 1 = (x − 2)

^{3}

We have:

x

^{3}− 4x

^{2}− x + 1 = (x − 2)

^{3}

⇒ x

^{3}− 4x

^{2}− x + 1 = x

^{3}+ 3x

^{2}(− 2) + 3x(− 2)

^{2}+ (− 2)

^{3}

⇒ x

^{3}− 4x

^{2}− x + 1 = x

^{3}− 6x

^{2}+ 12x − 8

⇒ x

^{3}− 4x

^{2}− x − 1 − x

^{3}+ 6x

^{2}− 12x + 8 = 0

⇒ 2x

^{2}− 13x + 9 = 0

Since 2x

^{2}− 13x + 9 is a quadratic polynomial

∴ x

^{3}− 4x

^{2}− x + 1 = (x − 2)

^{3}is a quadratic equation.

**Represent the following situations in the form of quadratic equations:**

**Question. The area of a rectangular plot is 528 m ^{2}. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.**

**Answer :**Let the breadth = x metres

ä Length = 2 (Breadth) + 1

∴ Length = (2x + 1) metres

Since Length × Breadth = Area

∴ (2x + 1) × x = 528

⇒ 2x

^{2}+ x = 528

⇒ 2x

^{2}+ x − 528 = 0

Thus, the required quadratic equation is

2x

^{2}+ x − 528 = 0

**Question. The product of two consecutive positive integers is 306. We need to find the integers.****Answer : **Let the two consecutive numbers be x and (x + 1).

QProduct of the numbers = 306

∴ x (x + 1) = 306

⇒ x^{2} + x = 306

⇒ x^{2} + x − 306 = 0

Thus, the required equdratic equation is

x^{2} + x − 306 = 0

**Question. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.****Answer :** Let the present age = x

∴ Mother’s age = (x + 26) years

After 3 years

His age = (x + 3) years

Mother’s age = [(x + 26) + 3] years

= (x + 29) years

According to the condition,

Product of their ages after 3 years = 360

⇒ (x + 3) × (x + 29) = 360

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 29x + 3x + 87 − 360 = 0

⇒ x^{2} + 32x − 273 = 0

Thus, the required quadratic equation is

x^{2} + 32x − 273 = 0

**Question. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.****Answer :** Let the speed of the train = u km/hr

Distance covered = 480 km

Time taken = Distance ÷ Speed

= (480 ÷ u) hours

= 480/u hours

In second case,

Speed = (u − 8) km/hour

∴ Time taken = Distance/speed = 480/u − 8 hours

According to the condition,

480/u − 8 u − 8 480/u hour

⇒ 480u − 480(u − 8) = 3u(u − 8)

⇒ 480u − 480u + 3840 = 3u^{2} − 24u

⇒ 3840 − 3u^{2} + 24u = 0

⇒ 1280 − u^{2} + 8u = 0

⇒ − 1280 + u^{2} − 8u = 0

⇒ u^{2} − 8u − 1280 = 0

Thus, the required quadratic equation is

u^{2} − 8u – 1280 = 0

**Find the roots of the following quadratic equations by factorisation:**

**Question. x ^{2} − 3x − 10 = 0**

**Answer :**x

^{2}− 3x − 10 = 0

We have:

x

^{2}− 3x − 10 = 0

⇒ x

^{2}− 5x + 2x − 10 = 0

⇒ x (x − 5) + 2 (x − 5) = 0

⇒ (x − 5) (x + 2) = 0

Either x − 5 = 0 ⇒ x = 5

or x + 2 = 0 ⇒ x = − 2

Thus, the required roots are x = 5 and x = −2.

**Question. 2x ^{2} + x − 6 = 0**

**Answer :**2x

^{2}+ x − 6 = 0

We have:

2x

^{2}+ x − 6 = 0

⇒ 2x

^{2}+ 4x − 3x − 6 = 0

⇒ 2x(x + 2) − 3 (x + 2) = 0

⇒ (x + 2) (2x − 3) = 0

Either x + 2 = 0 ⇒ x = −2

or 2x − 3 = 0 ⇒ x = 3/2

Thus, the required roots are x = −2 and x = 3/2 .

**Question. √2x ^{2} + 7x + 5√2 = 0**

**Answer :**√2x

^{2}+ 7x + 5√2 = 0

We have:

√2x

^{2}+ 7x + 5√2 = 0

⇒ √2x

^{2}+ 2x + 5x + 5√2 = 0

⇒ √2x

^{2}+ (√2⋅√2) x + 5x + 5.√2 = 0

⇒ √2x [x + √2] + 5 [x + √2] = 0

⇒ (x + √2) (√2 + 5) = 0

Either x + √2 = 0 ⇒ x = −√2

or √2x + 5 = 0 ⇒ x = − 5/√2

Thus, the required roots are x = −√2 and x = − 5/√2.

**Question. 2x ^{2} −x + 1/8 = 0**

**Answer :**2x

^{2}− x +x

^{2}

We have:

2x

^{2}− x + x

^{2}

⇒ 16x

^{2}− 8x + 1 = 0

⇒ 16x

^{2}− 4x − 4x + 1 = 0

⇒ 4x(4x − 1) − 1(4x − 1) = 0

⇒ (4x − 1) (4x − 1) = 0

⇒ x = 1/4 and x = 1/4

Thus, the required roots are x = 1/4 and x = 1/4.

**Question. 100x ^{2} − 20x + 1 = 0**

**Answer :**100x

^{2}− 20x + 1 = 0

We have:

100x

^{2}− 20x + 1 = 0

⇒ 100x

^{2}− 10x − 10x + 1 = 0

⇒ 10x (10x − 1) − 1 (10x − 1) = 0

⇒ (10x − 1) (10x − 1) = 0

⇒ (10x − 1) = 0 and (10x − 1) = 0

⇒ x = 1/10 and x = 1/10

Thus, the required roots are x = 1/10 and x = 1/10.

**Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:**

**Question. 2x ^{2} − 3x + 5 = 0**

**Answer :**2x

^{2}− 3x + 5 = 0

Comparing the given quadratic equation with ax

^{2}+ bx + c = 0, we have:

a = 2

b = − 3

c = 5

∴ The discriminant = b

^{2}− 4ac

= (− 3)

^{2}− 4 (2) (5)

= 9 − 40

= − 31 < 0

Since b

^{2}− 4ac is negative.

∴ The given quadratic equation has no real roots.

**Question. 3x ^{2} − 4√3 x + 4 = 0**

**Answer :**3x

^{2}− 4√3 x + 4 = 0

Comparing the given quadratic equation with ax

^{2}+ bx + c = 0, we get

a = 3

b = −4√3

c = 4

∴ b

^{2}− 4ac = [− 4√3] − 4(3)(4)

2 a fa f

= (16 × 3) − 48

= 48 − 48 = 0

Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:

− b/2a and − b/2a

i.e., −(−4√3)/2 × 3 and −(−4√3)/2 × 3

⇒ 4√3/2√3 × √3 and 4√3/2√3 × √3

⇒ 2/√3 and 2/√3

Thus, x = 2/√3 and x = 2/√3

**Question. 2x ^{2} − 6x + 3 = 0**

**Answer :**2x

^{2}− 6x + 3 = 0

Comparing it with the general quadratic equation, we have:

a = 2

b = − 6

c = 3

∴ b

^{2}− 4ac = (− 6)

^{2}− 4 (2) (3)

= 36 − 24

= 12 > 0

∴ The given quadratic equation has two real and distinct roots, which are given by

x = − b ± √b

^{2}− 4ac/2a

⇒ x = − (−6) ± √12/2 × 2

= 6 ± 2 √3/4

= 3 ± √3/2

Thus, the roots are:

x = 3 + √3/2 and x = 3 − √3/2

**Find the values of k for each of the following quadratic equations, so that they have two equal roots:**

**Question. 2x ^{2} + kx + 3 = 0**

**Answer**

**:**2x

^{2}+ kx + 3 = 0

Comparing the given quadratic equation with ax

^{2}+ bx + c = 0, we get

a = 2

b = k

c = 3

∴ b

^{2}− 4ac = (− k)

^{2}− 4 (2) (3)

= k

^{2}− 24

Q For a quadratic equation to have equal roots,

b

^{2}− 4ac = 0

∴ k

^{2}− 24 = 0 ⇒ k = ± √24

⇒ k = ±2√6

Thus, the required values of k are

2√6 and −2√6

**Question. kx (x − 2) + 6 = 0****Answer :** kx (x − 2) + 6 = 0

Comparing kx (x − 2) + 6 = 0 i.e., kx^{2} − 2kx + 6 = 0 with ax^{2} + bx + c = 0, we get

a = k

b = − 2k

c = 6

∴ b^{2} − 4ac = (− 2k)^{2} − 4 (k) (6)

= 4k^{2} − 24k

Since, the roots are real and equal,

∴ b^{2} − 4ac = 0

⇒ 4k^{2} − 24k = 0

⇒ 4k (k − 6) = 0

⇒ 4k = 0 or k − 6 = 0

⇒ k = 0 or k = 6

But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.

**Question. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m? If so, find its length and breadth.****Answer :** Let the breadth be x metres.

∴ Length = 2x metres

Now, Area = Length × Breadth

= 2x × x metre^{2}

= 2x^{2} sq. metre.

According to the given condition,

2x^{2} = 800

⇒ x^{2} = 800/2 = 400

⇒ x = ± √400 = ± 20

Therefore, x = 20 and x = − 20

But x = − 20 is possible (∴ breadth cannot be negative).

∴ x = 20

⇒ 2x = 2 × 20 = 40

Thus, length = 40 m and breadth = 20 m

**Question. Is the following situation possible? If so, determine their present ages.****The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.****Answer :** Let the age of one friend = x years

∴ The age of the other friend = (20 − x) years [∴ Sum of their ages is 20 years]

Four years ago

Age of one friend = (x − 4) years

Age of the other friend = (20 − x − 4) years

= (16 − x) years

According to the condition,

(x − 4) × (16 − x) = 48

⇒ 16x − 64 − x^{2} − 4x = 48

⇒ − x^{2} − 20x − 64 − 48 = 0

⇒ − x^{2} − 20x − 112 = 0

⇒ x^{2} + 20x + 112 = 0

Here, a = 1, b = 20 and c = 112

∴ b^{2} − 4ac = (20)^{2} − 4 (1) (112)

= 400 − 448

= − 48 < 0

Since b^{2} − 4ac is less than 0.

∴ The quadratic equation (1) has no real roots.

Thus, the given equation is not possible.

**Question. Is it possible to design a rectangular park of perimeter 80 m and area 400 m ^{2}? If so, find its length and breadth.**

**Answer :**Let the breadth of the rectangle be x metres.

Since, the parimetre of the rectangle = 80 m.

∴ 2 [Length + Breadth] = 80

2 [Length + x] = 80

⇒ Length + x = 80/2 = 40

⇒ Length = (40 − x) metres

∴ Area of the rectangle = Length × breadth

= (40 − x) × x sq. m

= 40x − x2

Now, according to the given condition,

Area of the rectangle = 400 m

^{2}

∴ 40x − x

^{2}= 400

⇒ − x

^{2}+ 40x − 400 = 0

⇒ x

^{2}− 40x + 400 = 0

Comparing (1) with ax

^{2}+ bx + c = 0, we get

a = 1

b = − 40

c = 400

∴ b

^{2}− 4ac = (− 40)

^{2}− 4 (1) (400)

= 1600 − 1600 = 0

Thus, the equation (1) has two equal and real roots.

∴ x = − b/2a and x = − b/2a

∴ breadth = − (−40)/2(1) = 40/2 = 20

∴ Breadth, x = 20 m

∴ Length = (40 − x) = (40 − 20) m = 20 m.

**Please click on below link to download CBSE Class 10 Mathematics Quadratic Equations Worksheet Set B**

## Books recommended by teachers

CBSE Class 10 Mathematics Probability And Constructions Worksheet Set A |

CBSE Class 10 Maths Probabilty Worksheet |