An equation with one variable, in which the highest power of the variable is two, is known as quadratic equation. Students who wish to develop a strong grip over the concepts of Selina can use Selina Solutions for this purpose. This resource can also be used by students for preparation of their exams too. All the solutions are prepared by subject experts at BYJUâ€™S to match all levels of students.

Chapter 5 of Concise Selina for Class 10 Maths has six exercises. Understanding quadratic equations, examining the nature of the roots, solving quadratic equations by factorisation and quadratic formula are the key topics covered in this chapter. Also, students can make use of the Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations PDF from the links given below.

## Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Download PDF

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 5 Quadratic Equations

## Access Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations

Exercise 5(A) Page No: 54

**1. Find which of the following equations are quadratic: **

**(i) (3x – 1) ^{2} = 5(x + 8)**

**(ii) 5x ^{2} â€“ 8x = -3(7 â€“ 2x)**

**(iii) (x -4) (3x + 1) = (3x â€“ 1) (x + 2)**

**(iv) x ^{2} + 5x â€“ 5 = (x – 3)^{2}**

**(v) 7x ^{3} â€“ 2x^{2} + 10 = (2x – 5)^{2} **

**(vi) (x – 1) ^{2} + (x + 2)^{2} + 3(x + 1) = 0**

**Solution: **

(i) (3x – 1)^{2}Â = 5(x + 8)

â‡’Â (9x^{2}Â – 6x + 1) = 5x + 40

â‡’Â 9x^{2}Â – 11x – 39 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

(ii) 5x^{2}Â – 8x = -3(7 – 2x)

â‡’Â 5x^{2}Â – 8x = 6x – 21

â‡’Â 5x^{2}Â – 14x + 21 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

(iii) (x – 4) (3x + 1) = (3x – 1) (x +2)

â‡’Â 3x^{2}Â + x – 12x – 4 = 3x^{2}Â + 6x – x – 2

â‡’Â 16x + 2 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a linear equation.

Thus, the given equation is not a quadratic equation.

(iv) x^{2}Â + 5x – 5 = (x – 3)^{2}

â‡’Â x^{2}Â + 5x – 5 = x^{2}Â – 6x + 9

â‡’Â 11x – 14 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a linear equation.

Thus, the given equation is not a quadratic equation.

(v) 7x^{3}Â – 2x^{2}Â + 10 = (2x – 5)^{2}

â‡’Â 7x^{3}Â – 2x^{2}Â + 10 = 4x^{2}Â – 20x + 25

â‡’Â 7x^{3}Â – 6x^{2}Â + 20x – 15 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a cubic equation.

Thus, the given equation is not a quadratic equation.

(vi) (x – 1)^{2}Â + (x + 2)^{2}Â + 3(x +1) = 0

â‡’Â x^{2}Â – 2x + 1 + x^{2}Â + 4x + 4 + 3x + 3 = 0

â‡’Â 2x^{2}Â + 5x + 8 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

**2. (i) Is x = 5 a solution of the quadratic equation x ^{2}Â – 2x – 15 = 0?**

**Solution: **

Given quadratic equation, x^{2}Â – 2x – 15 = 0

We know that, for x = 5 to be a solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = (5)^{2}Â – 2(5) – 15

= 25 – 10 – 15

= 0

= R.H.S

Therefore, x = 5 is a solution of the given quadratic equation x^{2}Â – 2x – 15 = 0

**(ii) Is x = -3 a solution of the quadratic equation 2x ^{2}Â – 7x + 9 = 0?**

**Solution: **

Given quadratic equation, 2x^{2}Â – 7x + 9 = 0

We know that, for x = -3 to be solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = 2(-3)^{2}Â – 7(-3) + 9

= 18 + 21 + 9

=Â 48

â‰ Â R.H.S

Therefore, x = -3 is not a solution of the given quadratic equation 2x^{2}Â – 7x + 9 = 0.

Exercise 5(B) Page No: 56

**1. Without solving, comment upon the nature of roots of each of the following equations: **

**(i) 7x ^{2}Â – 9x +2 = 0 (ii) 6x^{2}Â – 13x +4 = 0**

**(iii) 25x ^{2}Â – 10x + 1= 0 (iv) x^{2} + 2âˆš3x â€“ 9 = 0**

**(v) x ^{2}Â – ax – b^{2}Â = 0 (vi) 2x^{2}Â + 8x + 9 = 0**

**Solution: **

(i) Given quadratic equation, 7x^{2}Â – 9x + 2 = 0

Here, a = 7, b = -9 and c = 2

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (-9)^{2} â€“ 4(7)(2)

= 81 â€“ 56

= 25

As D > 0, the roots of the equation is real and unequal.

(ii) Given quadratic equation, 6x^{2}Â – 13x + 4 = 0

Here, a = 6, b = -13 and c = 4

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (-13)^{2} â€“ 4(6)(4)

= 169 â€“ 48

= 121

As D > 0, the roots of the equation is real and unequal.

(iii) Given quadratic equation, 25x^{2}Â – 10x + 1 = 0

Here, a = 25, b = -10 and c = 1

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (-10)^{2} â€“ 4(25)(1)

= 100 â€“ 100

= 0

As D = 0, the roots of the equation is real and equal.

(iv) Given quadratic equation, x^{2} + 2âˆš3x â€“ 9 = 0

Here, a = 1, b = 2âˆš3 and c = -9

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (2âˆš3)^{2} â€“ 4(1)(-9)

= 12 + 36

= 48

As D > 0, the roots of the equation is real and unequal.

(v) Given quadratic equation, x^{2}Â – ax – b^{2} = 0

Here, a = 1, b = -a and c = -b^{2}

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (a)^{2} â€“ 4(1)(-b^{2})

= a^{2} + 4b^{2}

a^{2} + 4b^{2} is always positive value.

Thus D > 0, and the roots of the equation is real and unequal

(vi) Given quadratic equation, 2x^{2}Â + 8x + 9 = 0

Here, a = 2, b = 8 and c = 9

So, the Discriminant (D) = b^{2} â€“ 4ac

D = (8)^{2} â€“ 4(2)(9)

= 64 â€“ 72

= -8

As D < 0, the equation has no roots.

**2. Find the value of â€˜pâ€™, if the following quadratic equations has equal roots: **

**(i) 4x ^{2}Â – (p – 2)x + 1 = 0**

**(ii) x ^{2} + (p – 3)x + p = 0**

**Solution:**

(i) 4x^{2}Â – (p – 2)x + 1 = 0

Here, a = 4, b = -(p – 2), c = 1

Given that the roots are equal,

So, Discriminant = 0 â‡’Â b^{2}– 4ac = 0

D =Â (-(p – 2))^{2 }– 4(4)(1) = 0

â‡’ p^{2}Â + 4 – 4p – 16 = 0

â‡’ p^{2} – 4p – 12 = 0

â‡’ p^{2} – 6p + 2p – 12 = 0

â‡’ p(p â€“ 6) + 2(p â€“ 6) = 0

â‡’ (p + 2)(p – 6) = 0

â‡’ p + 2 = 0 or p – 6 = 0

Hence, p = -2 or p = 6

(ii) x^{2}Â + (p – 3)x + p = 0

Here, a = 1, b = (p – 3), c = p

Given that the roots are equal,

So, Discriminant = 0 â‡’Â b^{2}– 4ac = 0

D =Â (p – 3)^{2 }– 4(1)(p) = 0

â‡’ p^{2}Â + 9 – 6p – 4p = 0

â‡’Â p^{2}– 10p + 9 = 0

â‡’ p^{2}-9p – p + 9 = 0

â‡’ p(p – 9) – 1(p – 9) = 0

â‡’Â (p -9)(p – 1) = 0

â‡’Â p – 9 = 0 or p – 1 = 0

Hence,Â p = 9 or p = 1

Exercise 5(C) Page No: 59

**Solve equations, number 1 to 20, given below, using factorization method: **

**1. x ^{2} â€“ 10x â€“ 24 = 0**

**Solution: **

Given equation, x^{2} â€“ 10x â€“ 24 = 0

x^{2} â€“ 12x + 2x â€“ 24 = 0

x(x – 12) + 2(x – 12) = 0

(x + 2)(x – 12) = 0

So, x + 2 = 0 or x – 12 = 0

Hence,

x = -2 or x = 12

**2. x ^{2} â€“ 16 = 0**

**Solution: **

Given equation, x^{2} â€“ 16 = 0

x^{2} + 4x â€“ 4x + 16 = 0

x(x + 4) -4(x + 4) = 0

(x – 4) (x + 4) = 0

So, (x – 4) = 0 or (x + 4) = 0

Hence,

x = 4 or x = -4

**3. 2x ^{2} â€“ Â½ x = 0 **

**Solution: **

Given equation, 2x^{2} â€“ Â½ x = 0

4x^{2} â€“ x = 0

x(4x – 1) = 0

So, either x = 0 or 4x â€“ 1 = 0

Hence,

x = 0 or x = Â¼

**4. x(x – 5) = 24**

**Solution: **

Given equation, x(x – 5) = 24

x^{2} â€“ 5x = 24

x^{2} â€“ 5x â€“ 24 = 0

x^{2} â€“ 8x + 3x â€“ 24 = 0

x(x – 8) + 3(x – 8) = 0

(x + 3)(x – 8) = 0

So, x + 3 = 0 or x – 8 = 0

Hence,

x = -3 or x = 8

**5. 9/2 x = 5 + x ^{2}**

**Solution: **

Given equation, 9/2 x = 5 + x^{2}

On multiplying by 2 both sides, we have

9x = 2(5 + x^{2})

9x = 10 + 2x^{2}

2x^{2} – 9x + 10 = 0

2x^{2} – 4x â€“ 5x + 10 = 0

2x(x – 2) â€“ 5(x – 2) = 0

(2x – 5)(x -2) = 0

So, 2x â€“ 5 = 0 or x â€“ 2 = 0

Hence,

x = 5/2 or x = 2

**6. 6/x = 1 + x **

**Solution: **

Given equation, 6/x = 1 + x

On multiplying by x both sides, we have

6 = x(1 + x)

6 = x + x^{2}

x^{2} + x â€“ 6 = 0

x^{2} + 3x â€“ 2x â€“ 6 = 0

x(x + 3) â€“ 2(x + 3) = 0

(x – 2) (x + 3) = 0

So, x â€“ 2 = 0 or x + 3 = 0

Hence,

x = 2 or x = -3

**7. x = (3x + 1)/ 4x **

**Solution: **

Given equation, x = (3x + 1)/ 4x

On multiplying by 4x both sides, we have

4x(x) = 3x + 1

4x^{2} = 3x + 1

4x^{2} â€“ 3x â€“ 1 = 0

4x^{2} â€“ 4x + x â€“ 1 = 0

4x(x – 1) + 1(x – 1) = 0

(4x + 1) (x – 1) = 0

So, 4x + 1 = 0 or x â€“ 1 = 0

Hence,

x = -1/4 or x = 1

**8. x + 1/x = 2.5 **

**Solution: **

Given equation, x + 1/x = 2.5

x + 1/x = 5/2

Taking LCM on L.H.S, we have

(x^{2} + 1)/ x = 5/2

2(x^{2} + 1) = 5x

2x^{2} + 2 = 5x

2x^{2} – 5x + 2 = 0

2x^{2} – 4x – x + 2 = 0

2x(x – 2) -1(x – 2) = 0

(2x – 1)(x – 2) = 0

So, 2x – 1 = 0 or x â€“ 2 = 0

Hence,

x = Â½ or x = 2

**9. (2x – 3) ^{2} = 49 **

**Solution:**

Given equation, (2x – 3)^{2} = 49

Expanding the L.H.S, we have

4x^{2} â€“ 12x + 9 = 49

4x^{2} â€“ 12x â€“ 40 = 0

Dividing by 4 on both side

x^{2} â€“ 3x â€“ 10 = 0

x^{2} â€“ 5x + 2x â€“ 10 = 0

x(x – 5) + 2(x – 5) = 0

(x + 2) (x – 5) = 0

So, x + 2 = 0 or x â€“ 5 = 0

Hence,

x = -2 or 5

**10. 2(x ^{2} â€“ 6) = 3(x – 4)**

**Solution: **

Given equation, 2(x^{2} – 6) = 3(x – 4)

2x^{2} â€“ 12 = 3x â€“ 12

2x^{2} = 3x

x(2x – 3) = 0

So, x = 0 or (2x – 3) = 0

Hence,

x = 0 or x = 3/2

**11. (x + 1) (2x + 8) = (x + 7) (x + 3)**

**Solution: **

Given equation, (x + 1) (2x + 8) = (x + 7) (x + 3)

2x^{2} + 2x + 8x + 8 = x^{2} + 7x + 3x + 21

2x^{2} + 10x + 8 = x^{2} + 10x + 21

x^{2 } = 21 â€“ 8

x^{2 }â€“ 13 = 0

(x – âˆš13) (x + âˆš13) = 0

So, x – âˆš13 = 0 or x + âˆš13 = 0

Hence,

x = – âˆš13 or x = âˆš13

**12. x ^{2} â€“ (a + b)x + ab = 0**

**Solution: **

Given equation, x^{2} â€“ (a + b)x + ab = 0

x^{2} â€“ ax – bx + ab = 0

x(x – a) â€“ b(x – a) = 0

(x â€“ b) (x – a) = 0

So, x â€“ b = 0 or x â€“ a = 0

Hence,

x = b or x = a

**13. (x + 3) ^{2} â€“ 4(x + 3) â€“ 5 = 0**

**Solution: **

Given equation, (x + 3)^{2} â€“ 4(x + 3) â€“ 5 = 0

(x^{2} + 9 + 6x) â€“ 4x â€“ 12 â€“ 5 = 0

x^{2} + 2x â€“ 8 = 0

x^{2} + 4x â€“ 2x â€“ 8 = 0

x(x + 4) â€“ 2(x – 4) = 0

(x – 2)(x + 4) = 0

So, x â€“ 2 = 0 or x + 4 = 0

Hence,

x = 2 or x = -4

**14. 4(2x – 3) ^{2} â€“ (2x – 3) â€“ 14 = 0 **

**Solution:**

Given equation, 4(2x – 3)^{2} â€“ (2x – 3) â€“ 14 = 0

Let substitute 2x â€“ 3 = y

Then the equation becomes,

4y^{2} â€“ y â€“ 14 = 0

4y^{2} – 8y + 7y â€“ 14 = 0

4y(y – 2) + 7(y â€“ 2) = 0

(4y + 7)(y – 2) = 0

So, 4y + 7 = 0 or y – 2 = 0

Hence,

y = -7/4 or y = 2

But we have taken y = 2x â€“ 3

Thus,

2x â€“ 3 = -7/4 or 2x â€“ 3 = 2

2x = 5/ 4 or 2x = 5

x = 5/8 or x = 5/2

**15. 3x â€“ 2/ 2x- 3 = 3x â€“ 8/ x + 4**

**Solution: **

Given equation, 3x â€“ 2/ 2x- 3 = 3x â€“ 8/ x + 4

On cross-multiplying we have,

(3x – 2)(x + 4) = (3x â€“ 8)(2x – 3)

3x^{2} â€“ 2x + 12x â€“ 8 = 6x^{2} â€“ 16x â€“ 9x + 24

3x^{2 }+ 10x â€“ 8 = 6x^{2} â€“ 25x + 24

3x^{2} â€“ 35x + 32 = 0

3x^{2} â€“ 3x â€“ 32x + 32 = 0

3x(x – 1) â€“ 32(x – 1) = 0

(3x – 32)(x – 1) = 0

So, 3x â€“ 32 = 0 or x â€“ 1 = 0

Hence,

x = 32/3 or x = 1

**16. 2x ^{2} â€“ 9x + 10 = 0, when: **

**(i) x âˆˆ N (ii) x âˆˆ Q **

**Solution: **

Given equation, 2x^{2} â€“ 9x + 10 = 0

2x^{2} â€“ 4x â€“ 5x + 10 = 0

2x(x – 2) â€“ 5(x – 2) = 0

(2x – 5)(x – 2) = 0

So, 2x â€“ 5 = 0 or x – 2 = 0

Hence,

x = 5/2 or x = 2

(i) When x âˆˆ N

x = 2 is the solution.

(ii) When x âˆˆ Q

x = 2, 5/2 are the solutions

**17. **

**Solution: **

2(2x^{2} + 18) = 5(x^{2} â€“ 9)

4x^{2} + 36 = 5x^{2} â€“ 45

x^{2} â€“ 81 = 0

(x – 9)(x + 9) = 0

So, x â€“ 9 = 0 or x + 9 = 0

Hence,

x = 9 or x = -9

Exercise 5(D) Page No: 59

**1. Solve, each of the following equations, using the formula: **

**(i) x ^{2} – 6x = 27**

**Solution: **

Given equation, x^{2} â€“ 6x = 27

x^{2} â€“ 6x â€“ 27 = 0

Here, a = 1 , b = -6 and c = -27

By quadratic formula, we have

Therefore, x = 9 or -3

**(ii) x ^{2} â€“ 10x + 21 = 0**

**Solution:**

Given equation, x^{2} â€“ 10x + 21 = 0

Here, a = 1, b = -10 and c = 21

By quadratic formula, we have

Therefore, x = 7 or x = 3

**(iii) x ^{2} + 6x â€“ 10 = 0 **

**Solution: **

Given equation, x^{2} + 6x â€“ 10 = 0

Here, a = 1, b = 6 and c = -10

By quadratic formula, we have

Therefore, x = -3 + âˆš19 or x = -3 – âˆš19

**(iv) x ^{2} + 2x â€“ 6 = 0**

**Solution: **

Given equation, x^{2} + 2x â€“ 6 = 0

Here, a = 1, b = 2 and c = -6

By quadratic formula, we have

Therefore, x = -1 + âˆš7 or x = -1 – âˆš7

**(v) 3x ^{2} + 2x â€“ 1 = 0 **

**Solution: **

Given equation, 3x^{2} + 2x â€“ 1 = 0

Here, a = 3, b = 2 and c = -1

By quadratic formula, we have

Therefore, x = 1/3 or x = -1

**(vi) 2x ^{2} + 7x + 5 = 0 **

**Solution: **

Given equation, 2x^{2} + 7x + 5 = 0

Here, a = 2, b = 7 and c = 5

By quadratic formula, we have

Therefore, x = -1 or x = -5/2

**(vii) 2/3 x = -1/6 x ^{2} â€“ 1/3 **

**Solution:**

Given equation, 2/3 x = -1/6 x^{2} â€“ 1/3

1/6 x^{2} + 2/3 x + 1/3 = 0

Multiplying by 6 on both sides

x^{2} + 4x + 2 = 0

Here, a = 1, b = 4 and c = 2

By quadratic formula, we have

Therefore, x = -2 + âˆš2 or x = -2 – âˆš2

**(viii) 1/15 x ^{2} + 5/3 = 2/3 x**

**Solution: **

Given equation, 1/15 x^{2} + 5/3 = 2/3 x

1/15 x^{2} – 2/3 x + 5/3 = 0

Multiplying by 15 on both sides

x^{2} – 10x + 25 = 0

Here, a = 1, b = -10 and c = 25

By quadratic formula, we have

Therefore, x = 5 (equal roots)

**(ix) x ^{2} – 6 = 2 âˆš2 x**

**Solution:**

Given equation, x^{2} – 6 = 2 âˆš2 x

x^{2} – 2âˆš2 x â€“ 6 = 0

Here, a = 1, b = -2âˆš2 and c = -6

By quadratic formula, we have

Therefore, x = 3âˆš2 or x = -âˆš2

**(x) 4/x â€“ 3 = 5/ (2x + 3)**

**Solution: **

Given equation, 4/x â€“ 3 = 5/ (2x + 3)

(4 â€“ 3x)/ x = 5/ (2x + 3)

On cross multiplying, we have

(4 â€“ 3x)(2x + 3) = 5x

8x â€“ 6x^{2} + 12 â€“ 9x = 5x

6x^{2} + 6x â€“ 12 = 0

Dividing by 6, we get

x^{2 }+ x â€“ 2 = 0

Here, a = 1, b = 1 and c = -2

By quadratic formula, we have

Therefore, x = 1 or x = -2

**(xi) 2x + 3/ x + 3 = x + 4/ x + 2 **

**Solution: **

Given equation, 2x + 3/ x + 3 = x + 4/ x + 2

On cross-multiplying, we have

(2x + 3) (x + 2) = (x + 4) (x + 3)

2x^{2} + 4x + 3x + 6 = x^{2} + 3x + 4x + 12

2x^{2} + 7x + 6 = x^{2 }+ 7x + 12

x^{2}Â + 0x â€“ 6 = 0

Here, a = 1, b = 0 and c = -6

By quadratic formula, we have

Therefore, x = âˆš6 or x = -âˆš6

**(xii) âˆš6x ^{2} â€“ 4x – 2âˆš6 = 0**

**Solution: **

Given equation, âˆš6x^{2} â€“ 4x – 2âˆš6 = 0

Here, a = âˆš6, b = -4 and c = -2âˆš6

By quadratic formula, we have

Therefore, x = âˆš6 or -âˆš6/3

**(xiii) 2x/ x â€“ 4 + (2x – 5)/(x – 3) =**

**Solution: **

Given equation, 2x/ x â€“ 4 + (2x – 5)/(x – 3) =

25x^{2} â€“ 175x + 300 = 12x^{2} â€“ 57x + 60

13x^{2} â€“ 118x + 240 = 0

Here, a = 13, b = -118 and c = 240

By quadratic formula, we have

Therefore, x = 6 or x = 40/13

**(ix) **

**Solution: **

From the given equation,

10x^{2} â€“ 60x + 80 = 6x^{2}Â â€“ 30x + 30

4x^{2} â€“ 30x + 50 = 0

2x^{2} â€“ 15x + 25 = 0

Here, a = 2, b = -15 and c = 25

Therefore, x = 5 or x = 5/2

**2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:**

**(i) x ^{2}Â – 8x +5 = 0**

**(ii) 5x ^{2}Â + 10x – 3 = 0**

**Solution: **

(i) x^{2}Â – 8x + 5 = 0

Here, a = 1, b = -8 and c = 5

By quadratic formula, we have

x = 4 Â± 3.3

Thus, x = 7.7 or x = 0.7

(ii) 5x^{2} + 10x â€“ 3 = 0

Here, a = 5, b = 10 and c = -3

By quadratic formula, we have

Thus, x = 0.3 or x = -2.3

**3. Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places: **

**(i) 2x ^{2} â€“ 10x + 5 = 0 **

**Solution:**

Given equation, 2x^{2} â€“ 10x + 5 = 0

Here, a = 2, b = -10 and c = 5

Therefore, x = 4.44 or x = 0.56

**(ii) 4x + 6/x + 13 = 0 **

**Solution: **

Given equation, 4x + 6/x + 13 = 0

Multiplying by x both sides, we get

4x^{2} + 13x + 6 = 0

Here, a = 4, b = 13 and c = 6

Therefore, x = -0.56 or x = -2.70

**(iii) 4x ^{2} â€“ 5x â€“ 3 = 0 **

**Solution: **

Given equation, 4x^{2} â€“ 5x â€“ 3 = 0

Here, a = 4, b = -5 and c = -3

Therefore, x = 1.70 or x = -0.44

**(iv) x ^{2} â€“ 3x â€“ 9 = 0**

**Solution:**

Given equation, x^{2} â€“ 3x â€“ 9 = 0

Here, a = 1, b = -3 and c = -9

Therefore, x = 4.85 or x = -1.85

**(v) x ^{2 }â€“ 5x â€“ 10 = 0 **

**Solution:**

Given equation, x^{2 }â€“ 5x â€“ 10 = 0

Here, a = 1, b = -5 and c = -10

Therefore, x = 6.53 or x = -1.53

**4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:**

**(i) 3x ^{2}Â – 12x – 1 = 0**

**(ii) x ^{2}Â – 16 x +6 = 0**

**(iii) 2x ^{2}Â + 11x + 4 = 0**

**Solution: **

(i) Given equation, 3x^{2}Â – 12x – 1 = 0

Here, a = 3, b = -12 and c = -1

Therefore, x = 4.082 or x = -0.082

(ii) Given equation, x^{2}Â – 16 x + 6 = 0

Here, a = 1, b = -16 and c = 6

Therefore, x = 15.616 or x = 0.384

(iii) Given equation, 2x^{2}Â + 11x + 4 = 0

Here, a = 2, b = 11 and c = 4

Therefore, x = -0.392 or x = -5.110

**5. Solve: **

**(i) x ^{4} â€“ 2x^{2}Â â€“ 3 = 0 **

**Solution: **

Given equation, x^{4} â€“ 2x^{2}Â â€“ 3 = 0

x^{4} â€“ 3x^{2}Â + x^{2 }â€“ 3 = 0

x^{2}(x^{2} – 3) + 1(x^{2 }– 3) = 0

(x^{2} + 1) (x^{2} – 3) = 0

So, x^{2} + 1 = 0 (which is not possible) or x^{2} â€“ 3 = 0

Hence,

x^{2} â€“ 3 = 0

x = Â± âˆš3

**(ii) x ^{4} â€“ 10x^{2} + 9 = 0**

**Solution:**

Given equation, x^{4} â€“ 10x^{2} + 9 = 0

x^{4} â€“ x^{2} â€“ 9x^{2}+ 9 = 0

x^{2}(x^{2 }– 1) â€“ 9(x^{2} – 1) = 0

(x^{2} – 9)(x^{2} – 1) = 0

So, we have

x^{2} â€“ 9 = 0 or x^{2} â€“ 1 = 0

Hence,

x = Â± 3 or x = Â± 1

Exercise 5(E) Page No: 66

**1. Solve each of the following equations: **

**Solution: **

Given equation,

4x^{2} + 6x + x â€“ 3 + 3x + 9 = 0

4x^{2} + 10x + 6 = 0

4x^{2} + 4x + 6x + 6 = 0

4x(x + 1) + 6(x + 1) = 0

(4x + 6) (x + 1) = 0

So, 4x + 6 = 0 or x + 1 = 0

x = -1 or x = -6/4 = -3/2 (rejected as this value is excluded in the domain)

Therefore,

x = -1 is the only solution

**2. (2x + 3) ^{2} = 81 **

**Solution: **

Given, (2x + 3)^{2} = 81

Taking square root on both sides we have,

2x + 3 = Â± 9

2x = Â± 9 â€“ 3

x = (Â± 9 â€“ 3)/ 2

So,

x = (9 â€“ 3)/ 2 or (-9 â€“ 3)/ 2

Therefore,

x = 3 or x = -6

**3. a ^{2}x^{2} â€“ b^{2} = 0**

**Solution: **

Given equation, a^{2}x^{2} â€“ b^{2} = 0

(ax)^{2} â€“ b^{2} = 0

(ax + b)(ax – b) = 0

So,

ax + b = 0 or ax â€“ b = 0

Therefore,

x = -b/a or b/a

**4. x ^{2} â€“ 11/4 x + 15/8 = 0 **

**Solution:**

Given equation, x^{2} â€“ 11/4 x + 15/8 = 0

Taking L.C.M we have,

(8x^{2} â€“ 22x + 15)/ 8 = 0

8x^{2} â€“ 22x + 15 = 0

8x^{2} â€“ 12x â€“ 10x + 15 = 0

4x(2x – 3) â€“ 5(2x – 3) = 0

(4x – 5)(2x – 3) = 0

So, 4x â€“ 5 = 0 or 2x â€“ 3 = 0

Therefore,

x = 5/4 or x = 3/2

**5. x + 4/x = -4; x â‰ 0**

**Solution: **

Given equation, x + 4/x = -4

(x^{2 }+ 4)/ x = -4

x^{2} + 4 = -4x

x^{2} + 4x + 4 = 0

x^{2} + 2x + 2x + 4 = 0

x(x + 2) + 2(x + 2) = 0

(x + 2)(x + 2) = 0

(x + 2)^{2} = 0

Taking square – root we have,

x + 2 = 0

Therefore, x = -2

**6. 2x ^{4} â€“ 5x^{2}Â + 3 = 0 **

**Solution: **

Given equation, 2x^{4} â€“ 5x^{2}Â + 3 = 0

Letâ€™s take x^{2} = y

Then, the equation becomes

2y^{2} â€“ 5y + 3 = 0

2y^{2} â€“ 2y â€“ 3y + 3 = 0

2y(y – 1) – 3(y – 1) = 0

(2y – 3) (y – 1) = 0

So, 2y â€“ 3 = 0 or y â€“ 1 = 0

y = 3/2 or y = 1

And, we have taken y = x^{2}

Thus,

x^{2} = 3/2 or x^{2} = 1

x = Â± âˆš(3/2) or x = Â±1

**7. x ^{4} â€“ 2x^{2}Â â€“ 3 = 0**

**Solution: **

Given equation, x^{4} â€“ 2x^{2}Â â€“ 3 = 0

Letâ€™s take x^{2} = y

Then, the equation becomes

y^{2} â€“ 2y â€“ 3 = 0

y^{2} â€“ 3y + y â€“ 3 = 0

y(y – 3) + 1(y – 3) = 0

(y + 1)(y – 3) = 0

So, y + 1 = 0 or y â€“ 3 = 0

y = -1 or y = 3

And, we have taken y = x^{2}

Thus,

x^{2} = – 1(impossible, no real solution)

x^{2} = 3

x = Â± âˆš3

**8. **

**Solution: **

Let us take (x + 1/x) = y â€¦. (1)

Now, squaring it on both sides

(x + 1/x)^{2} = y^{2}

x^{2} + 1/x^{2} + 2 = y^{2}

So,

x^{2} + 1/x^{2} = y^{2} â€“ 2 â€¦.. (2)

Using (1) and (2) in the given equation, we have

9(y^{2} â€“ 2) â€“ 9(y) â€“ 52 = 0

9y^{2} â€“ 18 â€“ 9y â€“ 52 = 0

9y^{2} â€“ 9y â€“ 70 = 0

9y^{2} â€“ 30y + 21y â€“ 70 = 0

3y(3y – 10) + 7(3y – 10) = 0

(3y + 7)(3y – 10) = 0

So, 3y + 7 = 0 or 3y â€“ 10 = 0

y = -7/3 or y = 10/3

Now,

x + 1/x = -7/3 or x + 1/x = 10/3

(x^{2} + 1)/x = -7/3 or (x^{2} + 1)/x = 10/3

3x^{2} â€“ 10x + 3 = 0 or 3x^{2} + 7x + 3 = 0

3x^{2} â€“ 9x â€“ x + 3 = 0 or

3x(x – 3) â€“ 1(x – 3) = 0

(3x – 1)(x – 3) = 0

So, x = 1/3 or 3

**9. **

**Solution: **

Let us take (x + 1/x) = y â€¦. (1)

Now, squaring it on both sides

(x + 1/x)^{2} = y^{2}

x^{2} + 1/x^{2} + 2 = y^{2}

So,

x^{2} + 1/x^{2} = y^{2} â€“ 2 â€¦.. (2)

Using (1) and (2) in the given equation, we have

2(y^{2} â€“ 2) â€“ (y) = 11

2y^{2} â€“ 4 â€“ y = 11

2y^{2} â€“ y â€“ 15 = 0

2y^{2} â€“ 6y + 5y â€“ 15 = 0

2y(y – 3) + 5(y – 3) = 0

(2y + 5) (y – 3) = 0

So,

2y + 5 = 0 or y – 3 = 0

y = -5/2 or y = 3

Now,

x + 1/x = -5/2 or x + 1/x = 3

(x^{2} + 1)/x = -5/2 or (x^{2} + 1)/x = 3

2(x^{2} + 1) = -5x or x^{2} + 1 = 3x

2x^{2} + 5x + 2 = 0 or x^{2 }– 3x + 1 = 0

2x^{2} + 4x + x + 2 = 0 or

2x(x + 2) + 1(x + 2) = 0

(2x + 1)(x + 2) = 0

Hence, x = -1/2 or -2

**10. **

**Solution: **

Let us take (x – 1/x) = y â€¦. (1)

Now, squaring it on both sides

(x – 1/x)^{2} = y^{2}

x^{2} + 1/x^{2} – 2 = y^{2}

So,

x^{2} + 1/x^{2} = y^{2} + 2 â€¦.. (2)

Using (1) and (2) in the given equation, we have

(y^{2} + 2) â€“ 3(y) â€“ 2 = 0

y^{2}Â -3y = 0

y(y – 3) = 0

So, y = 0 or y â€“ 3 = 0

Now,

(x – 1/x) = 0 or (x – 1/x) = 3

x^{2} â€“ 1 = 0 or x^{2} â€“ 1 = 3x

x^{2} = 1 or x^{2} â€“ 3x â€“ 1 = 0

Therefore,

x = Â± 1 or

Exercise 5(F) Page No: 67

**1. Solve: **

**(i) (x + 5) (x – 5) = 24**

**Solution: **

Given equation, (x + 5) (x – 5) = 24

x^{2} â€“ 25 = 24

x^{2} = 49

Thus,

x = Â± 7

**(ii) 3x ^{2} – 2âˆš6x + 2 = 0**

**Solution: **

Given equation, 3x^{2} – 2âˆš6x + 2 = 0

3x^{2} – âˆš6x – âˆš6x + 2 = 0

âˆš3x(âˆš3x – âˆš2) – âˆš2(âˆš3x – âˆš2) = 0

(âˆš3x – âˆš2) (âˆš3x – âˆš2) = 0

So, âˆš3x – âˆš2 = 0 or âˆš3x – âˆš2 = 0

Therefore,

x = âˆš(2/3), âˆš(2/3) (equal roots)

**(iii) 3âˆš2x ^{2} â€“ 5x – âˆš2 = 0 **

**Solution: **

Given equation, 3âˆš2x^{2} â€“ 5x – âˆš2 = 0

3âˆš2x^{2} â€“ 6x + x – âˆš2 = 0

3âˆš2x(x – âˆš2) + 1(x – âˆš2) = 0

(3âˆš2x + 1) (x – âˆš2) = 0

So, 3âˆš2x + 1 = 0 or x – âˆš2 = 0

Therefore,

x = -1/ 3âˆš2 or x = âˆš2

**(iv) 2x â€“ 3 = âˆš(2x ^{2} â€“ 2x + 21)**

**Solution: **

Given equation, 2x â€“ 3 = âˆš(2x^{2} â€“ 2x + 21)

On squaring on both sides, we have

(2x â€“ 3)^{2} = 2x^{2} â€“ 2x + 21

4x^{2} + 9 â€“ 12x = 2x^{2} â€“ 2x + 21

2x^{2} â€“ 10x â€“ 12 = 0

Dividing by 2, we get

x^{2} â€“ 5x â€“ 6 = 0

x^{2} â€“ 6x + x â€“ 6 = 0

x(x – 6) â€“ 1(x – 6) = 0

(x – 1) (x – 6) = 0

So, x â€“ 1 = 0 or x â€“ 6 = 0

Thus, we get

x = 1 or x = 6

But, putting x = 1 the L.H.S become negative. And we know that the square root function always gives a positive vale.

Therefore,

x = 6 is the only solution.

**2. One root of the quadratic equationÂ 8x ^{2 }+ mx + 15 = 0 is Â¾. Find the value of m. Also, find the other root of the equation.Â **

**Solution:**

Given equation, 8x^{2 }+ mx + 15 = 0

One of the roots is Â¾, and hence it satisfies the given equation

So,

8(3/4)^{2 }+ m(3/4) + 15 = 0

8(9/16) + m(3/4) + 15 = 0

18/4 + 3m/4 + 15 = 0

Taking L.C.M, we have

(18 + 3m + 60)/4 = 0

18 + 3m + 60 = 0

3m = – 78

m = -26

Now, putting the value of m in the given equation, we get

8x^{2 }+ (-26)x + 15 = 0

8x^{2 }– 26x + 15 = 0

8x^{2} â€“ 20x â€“ 6x + 15 = 0

4x(2x – 5) â€“ 3(2x – 5) = 0

(4x â€“ 3) (2x – 5) = 0

So, 4x â€“ 3 = 0 or 2x â€“ 5 = 0

Therefore.

x = Â¾ or x = 5/2

**3. Show that one root of the quadratic equation x ^{2} + (3 â€“ 2a)x â€“ 6a = 0 is -3. Hence, find its other root.**

**Solution: **

Given quadratic equation, x^{2} + (3 â€“ 2a)x â€“ 6a = 0

Now, putting x = -3 we have

(-3)^{2} + (3 â€“ 2a)( -3) â€“ 6a = 0

9 â€“ 9 + 6a â€“ 6a = 0

0 = 0

Since, x = -3 satisfies the given equation -3 is one of the root of the quadratic equation.

x^{2} + (3 â€“ 2a)x â€“ 6a = 0

x^{2} + 3x â€“ 2ax â€“ 6a = 0

x(x + 3) â€“ 2a(x + 3) = 0

(x â€“ 2a) (x + 3) = 0

So, x â€“ 2a = 0 or x + 3 =0

x = 2a or x = -3

Hence, the other root is 2a.

**4. If p â€“ 15 = 0 and 2x ^{2} + px + 25 = 0: find the values of x. **

**Solution: **

Given equations, p â€“ 15 = 0 and 2x^{2} + px + 25 = 0

Thus, p = 15

Now, using p in the quadratic equation, we get

2x^{2} + (15)x + 25 = 0

2x^{2} + 10x + 5x + 25 = 0

2x(x + 5) + 5(x + 5) = 0

(2x + 5) (x + 5) = 0

So, 2x + 5 = 0 or x + 5 = 0

Hence,

x = -5/2 or x = -5

**5. Find the solution of the quadratic equation 2x ^{2} â€“ mx â€“ 25n = 0; if m + 5 = 0 and n â€“ 1 = 0.**

**Solution: **

Given,

m + 5 = 0 and n â€“ 1 = 0

so,

m = -5 and n = 1

Now, putting these values in the given quadratic equation 2x^{2} â€“ mx â€“ 25n = 0, we get

2x^{2} â€“ (-5)x â€“ 25(1) = 0

2x^{2} + 5x â€“ 25 = 0

2x^{2} + 10x â€“ 5x â€“ 25 = 0

2x(x + 5) -5(x + 5) = 0

(2x – 5) (x + 5) = 0

So, 2x â€“ 5 = 0 or x + 5 = 0

Hence,

x = 5/2 or x = -5

**6. If m and n are roots of the equation: 1/x â€“ 1/(x-2) = 3: where x â‰ 0 and x â‰ 2; find m x n. **

**Solution: **

Given equation, 1/x â€“ 1/(x-2) = 3

(x – 2 – x)/ (x(x – 2)) = 3

-2 = 3(x^{2} â€“ 2x)

3x^{2} â€“ 6x + 2 = 0

Solving by using quadratic formula, we get

And, since m and n are roots of the equation, we have

m = (âˆš3 + 1)/ âˆš3 n = (âˆš3 – 1)/ âˆš3

So,

m x n = (âˆš3 + 1)/ âˆš3 x (âˆš3 – 1)/ âˆš3 = [(âˆš3)^{2} â€“ 1]/ (âˆš3)^{2}

Thus,

m x n = 2/3

*The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.*