## Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations

**Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations**

### Quadratic Equations Exercise 5A – Selina Concise Mathematics Class 10 ICSE Solutions

Find which of the following equations are quadratic:

Solution 1(i)

(3x – 1)^{2} = 5(x + 8)

⇒ (9x^{2} – 6x + 1) = 5x + 40

⇒ 9x^{2} – 11x – 39 =0; which is of the form ax^{2} + bx + c = 0.

∴ Given equation is a quadratic equation.

Solution 1(ii)

5x^{2} – 8x = -3(7 – 2x)

⇒ 5x^{2} – 8x = 6x – 21

⇒ 5x^{2} – 14x + 21 =0; which is of the form ax^{2} + bx + c = 0.

∴ Given equation is a quadratic equation.

Solution 1(iii)

(x – 4)(3x + 1) = (3x – 1)(x +2)

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

⇒ 16x + 2 =0; which is not of the form ax^{2} + bx + c = 0.

∴ Given equation is not a quadratic equation.

Solution 1(iv)

x^{2} + 5x – 5 = (x – 3)^{2 }⇒ x^{2} + 5x – 5 = x^{2} – 6x + 9

⇒ 11x – 14 =0; which is not of the form ax^{2} + bx + c = 0.

∴ Given equation is not a quadratic equation.

Solution 1(v)

7x^{3} – 2x^{2} + 10 = (2x – 5)^{2 }⇒ 7x^{3} – 2x^{2} + 10 = 4x^{2} – 20x + 25

⇒ 7x^{3} – 6x^{2} + 20x – 15 = 0; which is not of the form ax^{2} + bx + c = 0.

∴ Given equation is not a quadratic equation.

Solution 1(vi)

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

⇒ x^{2} – 2x + 1 + x^{2} + 4x + 4 + 3x + 3 = 0

⇒ 2x^{2} + 5x + 8 = 0; which is of the form ax^{2} + bx + c = 0.

∴ Given equation is a quadratic equation.

Question 2(i)

Is x = 5 a solution of the quadratic equation x^{2} – 2x – 15 = 0?

Solution:

x^{2} – 2x – 15 = 0

For x = 5 to be solution of the given quadratic equation it should satisfy the equation.

So, substituting x = 5 in the given equation, we get

L.H.S = (5)^{2} – 2(5) – 15

= 25 – 10 – 15

= 0

= R.H.S

Hence, x = 5 is a solution of the quadratic equation x^{2} – 2x – 15 = 0.

Question 2(ii).

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

Solution:

2x^{2} – 7x + 9 = 0

For x = -3 to be solution of the given quadratic equation it should satisfy the equation

So, substituting x = 5 in the given equation, we get

L.H.S =2(-3)^{2} – 7(-3) + 9

= 18 + 21 + 9

= 48

≠ R.H.S

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

Question 3.

If \(\sqrt{\frac{2}{3}}\) is a solution of equation 3x^{2} + mx + 2 = 0, find the value of m.

Solution:

For x = \(\sqrt{\frac{2}{3}}\) to be solution of the given quadratic equation it should satisfy the equation

So, substituting x = \(\sqrt{\frac{2}{3}}\) in the given equation, we get

Question 4.

\(\frac{2}{3}\) and 1 are the solutions of equation mx^{2} + nx + 6 = 0. Find the values of m and n.

Solution:

For x = \(\frac{2}{3}\) and x = 1 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = \(\frac{2}{3}\) and x = 1 in the given equation, we get

Solving equations (1) and (2) simultaneously,

4m + 6n + 54 = 0 …..(1)

m + n + 6 = 0 ….(2)

(1) – (2) × 6

⇒ -2m + 18 = 0

⇒ m = 9

Substitute in (2)

⇒ n = -15

Question 5.

If 3 and -3 are the solutions of equation ax^{2} + bx – 9 = 0. Find the values of a and b.

Solution:

For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = 3 and x = -3 in the given equation, we get

Solving equations (1) and (2) simultaneously,

9a + 3b – 9 = 0 …(1)

9a – 3b – 9 = 0 …(2)

(1) + (2)

⇒ 18a – 18 = 0

⇒ a = 1

Substitute in (2)

⇒ b = 0

### Quadratic Equations Exercise 5B – Selina Concise Mathematics Class 10 ICSE Solutions

**Question 1.**

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

(i) 7x^{2} – 9x +2 =0

(ii) 6x^{2} – 13x +4 =0

(iii) 25x^{2} – 10x +1=0

(iv) x^{2} + 2√3x – 9=0

(v) x^{2} – ax – b^{2} =0

(vi) 2x^{2} +8x +9=0

**Solution:**

**Question 2.**

Find the value of p, if the following quadratic equation has equal roots : 4x^{2} – (p – 2)x + 1 = 0

**Solution:**

**Question 3.**

Find the value of ‘p’, if the following quadratic equations have equal roots : x^{2} + (p – 3)x + p = 0

**Solution:**

x^{2} + (p – 3)x + p = 0

Here, a = 1, b = (p – 3), c = p

Since, the roots are equal,

⇒ b^{2}– 4ac = 0

⇒ (p – 3)^{2}– 4(1)(p) = 0

⇒p^{2} + 9 – 6p – 4p = 0

⇒ p^{2}– 10p + 9 = 0

⇒p^{2}-9p – p + 9 = 0

⇒p(p – 9) – 1(p – 9) = 0

⇒ (p -9)(p – 1) = 0

⇒ p – 9 = 0 or p – 1 = 0

⇒ p = 9 or p = 1

**Question 4.**

The equation 3x^{2} – 12x + (n – 5)=0 has equal roots. Find the value of n.

**Solution:**

**Question 5.**

Find the value of m, if the following equation has equal roots : (m – 2)x^{2} – (5+m)x +16 =0

**Solution:**

**Question 6.**

Find the value of p for which the equation 3x^{2}– 6x + k = 0 has distinct and real roots.

**Solution:**

### Quadratic Equations Exercise 5C – Selina Concise Mathematics Class 10 ICSE Solutions

**Question 1.**

Solve : x² – 10x – 24 = 0

**Solution:**

**Question 2.**

Solve : x² – 16 = 0

**Solution:**

**Question 3.**

**Solution:**

**Question 4.**

Solve : x(x – 5) = 24

**Solution:**

**Question 5.**

**Solution:**

**Question 6.**

**Solution:**

**Question 7.**

**Solution:**

**Question 8.**

**Solution:**

**Question 9.**

Solve : (2x – 3)² = 49

**Solution:**

**Question 10.**

Solve : 2(x² – 6) = 3(x – 4)

**Solution:**

**Question 11.**

Solve : (x + 1)(2x + 8) = (x + 7)(x + 3)

**Solution:**

**Question 12.**

Solve : x² – (a + b)x + ab = 0

**Solution:**

**Question 13.**

(x + 3)² – 4(x + 3) – 5 = 0

**Solution:**

**Question 14.**

4(2x – 3)² – (2x – 3) – 14 = 0

**Solution:**

**Question 15.**

**Solution:**

**Question 16.**

2x^{2} – 9x + 10 = 0, When

(i) x∈ N

(ii) x∈ Q

**Solution:**

**Question 17.**

**Solution:**

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

**Question 20.**

**Solution:**

**Question 21.**

Find the quadratic equation, whose solution set is :

(i) {3, 5} (ii) {-2, 3}

**Solution:**

**Question 22.**

**Solution:**

**Question 23.**

**Solution:**

**Question 24.**

Find the value of x, if a + 1=0 and x^{2} + ax – 6 =0.

**Solution:**

**Question 25.**

Find the value of x, if a + 7=0; b + 10=0 and 12x^{2} = ax – b.

**Solution:**

If a + 7 =0, then a = -7

and b + 10 =0, then b = – 10

Put these values of a and b in the given equation

**Question 26.**

Use the substitution y= 2x +3 to solve for x, if 4(2x+3)^{2} – (2x+3) – 14 =0.

**Solution:**

**Question 27.**

Without solving the quadratic equation 6x^{2} – x – 2=0, find whether x = 2/3 is a solution of this equation or not.

**Solution:**

**Question 28.**

Determine whether x = -1 is a root of the equation x^{2} – 3x +2=0 or not.

**Solution:**

x^{2} – 3x +2=0

Put x = -1 in L.H.S.

L.H.S. = (-1)^{2 }– 3(-1) +2

= 1 +3 +2=6 ≠ R.H.S

Then x = -1 is not the solution of the given equation.

**Question 29.**

If x = 2/3 is a solution of the quadratic equation 7x^{2}+mx – 3=0; Find the value of m.

**Solution:**

**Question 30.**

If x = -3 and x = 2/3 are solutions of quadratic equation mx^{2 }+ 7x + n = 0, find the values of m and n.

**Solution:**

**Question 31.**

If quadratic equation x^{2} – (m + 1) x + 6=0 has one root as x =3; find the value of m and the root of the equation.

**Solution:**

**Question 32.**

Given that 2 is a root of the equation 3x² – p(x + 1) = 0 and that the equation px² – qx + 9 = 0 has equal roots, find the values of p and q.

**Solution:**

**Question 33.**

**Solution:**

**Question 34.**

**Solution:**

**Question 35. **If -1 and 3 are the roots of x

^{2}+ px + q = 0, find the values of p and q.

**Solution:**

### Quadratic Equations Exercise 5D – Selina Concise Mathematics Class 10 ICSE Solutions

**Question 1.**

**Solution:**

**Question 2.**

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

(i) x^{2} – 8x+5=0

(ii) 5x^{2} +10x – 3 =0

**Solution:**

**Question 3(i).**

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

(i) 2x^{2} – 10x +5=0

**Solution:**

**Question 3(ii).**

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

4x + 6/x + 13 = 0

**Solution:**

**Question 3(iii).**

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

x^{2} – 3x – 9 =0

**Solution:**

**Question 3(iv).**

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

x^{2} – 5x – 10 = 0

**Solution:**

**Question 4.**

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

(i) 3x^{2} – 12x – 1 =0

(ii) x^{2} – 16 x +6= 0

(iii) 2x^{2} + 11x + 4= 0

**Solution:**

**Question 5.**

Solve:

(i) x^{4} – 2x^{2} – 3 =0

(ii) x^{4} – 10x^{2} +9 =0

**Solution:**

**Question 6.**

Solve :

(i) (x^{2} – x)^{2} + 5(x^{2} – x)+ 4=0

(ii) (x^{2} – 3x)^{2} – 16(x^{2} – 3x) – 36 =0

**Solution:**

**Question 7.**

**Solution:**

**Question 8.**

**Solution:**

**Question 9.**

Solve the following equation and give your answer correct to 3 significant figures:

5x² – 3x – 4 = 0

**Solution:**

**Question 10.**

Solve for x using the quadratic formula. Write your answer correct to two significant figures.

(x – 1)^{2} – 3x + 4 = 0

**Solution:**

**Question 11.**

Solve the quadratic equation x² – 3 (x+3) = 0; Give your answer correct to two significant figures.

**Solution: **

### Quadratic Equations Exercise 5E – Selina Concise Mathematics Class 10 ICSE Solutions

**Question 1.**

**Solution:**

**Question 2.**

Solve: (2x+3)^{2}=81

**Solution:**

**Question 3.**

Solve: a²x² – b² = 0

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

**Solution:**

**Question 6.**

Solve: 2x^{4} – 5x² + 3 = 0

**Solution:**

**Question 7.**

Solve: x^{4} – 2x² – 3 = 0.

**Solution:**

**Question 8.**

**Solution:**

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

**Question 11.**

Solve : (x² + 5x + 4)(x² + 5x + 6) = 120

**Solution:**

**Question 12.**

Solve each of the following equations, giving answer upto two decimal places.

(i) x^{2} – 5x -10=0 (ii) 3x^{2} – x – 7 =0

**Solution:**

**Question 13.**

**Solution:**

**Question 14.**

Solve :

(i) x^{2} – 11x – 12 =0; when x ∈ N

(ii) x^{2} – 4x – 12 =0; when x ∈ I

(iii) 2x^{2} – 9x + 10 =0; when x ∈ Q

**Solution:**

**Question 15.**

Solve : (a + b)²x² – (a + b)x – 6 = 0; a + b ≠ 0

**Solution:**

**Question 16.**

**Solution:**

**Question 17.**

**Solution:**

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

**Question 20.**

Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.

x² + 2(m – 1)x + (m + 5) = 0

**Solution:**

### Quadratic Equations Exercise 5F – Selina Concise Mathematics Class 10 ICSE Solutions

Solution 1(i)

Given: (x + 5)(x – 5)=24

⇒ x^{2} – 5^{2} = 24 …. since (a – b)(a + b) = a^{2} – b^{2}

⇒ x^{2} – 25 = 24

⇒ x^{2} = 49

⇒ x = ± 7

Solution 1(ii)

Given: 3x^{2} – 2\(\sqrt{6}\)x + 2 = 0

Solution 1(iii)

Given: 3\(\sqrt{2}\)x^{2} – 5x – \(\sqrt{26}\) = 0

Question 2.

One root of the quadratic equation 8x^{2} + mx + 15 is 3/4. Find the value of m. Also, find the other root of the equation.

Solution:

Given quadratic equation is 8x^{2} + mx + 15 = 0 …. (i)

One of the roots of (i) is \(\frac{3}{4}\), so it satisfies (i)

So, the equation (i) becomes 8x^{2} – 26x + 15 = 0

⇒ 8x^{2} – 20x – 6x + 15 = 0

⇒ 4x(2x – 5) -3(2x – 5) = 0

⇒ (4x – 3)(2x – 5) = 0

⇒ x = \(\frac{3}{4}\) or x = \(\frac{5}{2}\)

⇒ x = \(\frac{3}{4}, \frac{5}{2}\)

Hence, the other root is \(\frac{5}{2}\)

Question 3.

One root of the quadratic equation x^{2} – 3x – 2ax – 6a = 0 is -3, find its other root.

Solution:

Given quadratic equation is …. (i)

One of the roots of (i) is -3, so it satisfies (i)

⇒ x^{2} – 3x – 2ax – 6a = 0

⇒ x(x + 3) – 2a(x + 3) = 0

⇒ (x – 2a)(x + 3) = 0

⇒ x = -3, 2a

Hence, the other root is 2a.

Question 4.

If p – 15 = 0 and 2x^{2} + 15x + 15 = 0;find the values of x.

Solution:

Given i.e p – 15 = 0 i.e. p = 15

So, the given quadratic equation becomes

2x^{2} + 15x + 15 = 0

⇒ 2x + 10x + 5x + 15 = 0

⇒ 2x(x + 5) + 5(x + 5)

⇒ (2x + 5)(x + 5) = 0

⇒ x = -5, \(-\frac{5}{2}\)

Hence, the values of x are -5 and \(-\frac{5}{2}\)

Question 5.

Find the solution of the equation 2x^{2} -mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.

Solution:

Given quadratic equation is 2x^{2} -mx – 25n = 0 ….. (i)

Also, given and m + 5 = 0 and n – 1 = 0

⇒ m = -5 and n = 1

So, the equation (i) becomes

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

⇒ 2x + 10x – 5x – 25 = 0

⇒ 2x(x + 5) -5(x + 5) = 0

⇒ (x + 5)(2x – 5) = 0

⇒ x = -5, \(\frac{5}{2}\)

Hence, the solution of given quadratic equation are x and \(\frac{5}{2}\)

Question 6.

If m and n are roots of the equation \(\frac{1}{x}-\frac{1}{x-2}=3\) where x ≠ 0 and x ≠ 2; find m × n.

Solution:

Given quadratic equation is \(\frac{1}{x}-\frac{1}{x-2}=3\)

Since, m and n are roots of the equation, we have

Question 7.

Solve, using formula :

x^{2} + x – (a + 2)(a + 1) = 0

Solution:

Given quadratic equation is x^{2} + x – (a + 2)(a + 1) = 0

Using quadratic formula,

Question 8.

Solve the quadratic equation 8x^{2} – 14x + 3 = 0

(i) When x ∈ I (integers)

(ii) When x ∈ Q (rational numbers)

Solution:

Given quadratic equation is 8x^{2} – 14x + 3 = 0

⇒ 8x^{2} – 12x – 2x + 3 = 0

⇒ 4x(2x – 3) – (2x – 3) = 0

⇒ (4x – 1)(2x – 3) = 0

⇒ x = \(\frac{3}{2}\) or x = \(\frac{1}{4}\)

(i) When x ϵ I, the equation 8x^{2} – 14x + 3 = 0 has no roots

(ii) When x ϵ Q the roots of 8x^{2} – 14x + 3 = 0 are

x = \(\frac{3}{2}\) x = \(\frac{1}{4}\)

Question 9.

Find the value of m for which the equation (m + 4 )^{2} + (m + 1)x + 1 = 0 has real and equal roots.

Solution:

Given quadratic equation is (m + 4 )^{2} + (m + 1)x + 1 = 0

The quadratic equation has real and equal roots if its discriminant is zero.

⇒ D = b^{2} – 4ac = 0

⇒ (m + 1)^{2} -4(m + 4)(1) = 0

⇒ m^{2} + 2m + 1 – 4m – 16 = 0

⇒ m^{2} – 2m – 15 = 0

⇒ m^{2} – 5m + 3m – 15 = 0

⇒ m(m – 5) +3(m =5) = 0

⇒ (m – 5)(m + 3) = 0

⇒ m = 5 or m = -3

Question 10.

Find the values of m for which equation 3x^{2} + mx + 2 = 0 has equal roots. Also, find the roots of the given equation.

Solution:

Given quadratic equation is 3x^{2} + mx + 2 = 0 …. (i)

The quadratic equation has equal roots if its discriminant is zero

⇒ D = b^{2} – 4ac = 0

⇒ m^{2} – 4(2)(3) = 0

⇒ m^{2} = 24

⇒ m = \(\pm 2 \sqrt{6}\)

When m = \(2 \sqrt{6}\), equation (i) becomes

When m = \(-2 \sqrt{6}\), equation (i) becomes

∴ x= \(-\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3}\)

Question 11.

Find the value of k for which equation 4x^{2} + 8x – k = 0 has real roots.

Solution:

Given quadratic equation is 4x^{2} + 8x – k = 0 …. (i)

The quadratic equation has real roots if its discriminant is greater than or equal to zero

⇒ D = b^{2} – 4ac ≥ 0

⇒ 8^{2} – 4(4)(-k) ≥ 0

⇒ 64 + 16k ≥ 0

⇒ 16k ≥ -64

⇒ k ≥ -4

Hence, the given quadratic equation has real roots for k ≥ -4

Question 12.

Find, using quadratic formula, the roots of the following quadratic equations, if they exist

(i) 3x^{2} – 5x + 2 = 0

(ii) x^{2} + 4x + 5 = 0

Solution:

(i) Given quadratic equation is 3x^{2} – 5x + 2 = 0

D = b^{2} – 4ac = (-5)^{2} – 4(3)(2) = 25 – 24 = 1

Since D > 0, the roots of the given quadratic equation are real and distinct.

Using quadratic formula, we have

⇒ x = 1 or x = \(\frac{2}{3}\)

(ii) Given quadratic equation is x^{2} + 4x + 5 = 0

D = b^{2} – 4ac = (4)^{2} – 4(1)(5) = 16 – 20 = – 4

Since D < 0, the roots of the given quadratic equation does not exist.

Solution 13:

(i) Given quadratic equation is \(\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}\)

⇒ 48x = 324 – x^{2}

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + 54x – 6x – 324 = 0

⇒ x(x + 54) -6(x + 54) = 0

⇒ (x + 54)(x – 6) = 0

⇒ x = -54 or x = 6

But as x > 0, so x can’t be negative.

Hence, x = 6.

(ii) Given quadratic equation is \((x-10)\left(\frac{1200}{x}+2\right)=1260\)

⇒ (x – 10)\(\left(\frac{1200+2 x}{x}\right)\) = 1260

⇒ (x – 10)(1200 + 2x) = 1260x

⇒ 1200x + 2x^{2} – 12000 – 20x = 1260x

⇒ 2x^{2} – 12000 – 80x = 0

⇒ x^{2} – 40x – 6000 = 0

⇒ x^{2} – 100x + 60x – 6000 = 0

⇒ (x – 100)(x – 60) = 0

⇒ x = 100 or x = -60

But as x < 0, so x can’t be positive.

Hence, x = -60.

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