## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D.

**Other Exercises**

- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5A
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5B
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5C
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

**Solve each of the following equations :**

**Question 1.**

**Solution:**

**Question 2.**

**(2x + 3)² = 81**

**Solution:**

(2x + 3)2 = 81

⇒ 4x² + 12x + 9 = 81

⇒ 4x² + 12x + 9 – 81 = 0

⇒ 4x² + 12x – 72 = 0

⇒ x² + 3x – 18 = 0 (Dividing by 4)

⇒ x² + 6x – 3x – 18 = 0

⇒ x (x + 6) – 3 (x + 6) = 0

⇒ (x + 6) (x – 3) = 0

Either x + 6 = 0, then x = -6

or x – 3 = 0, then x = 3

x = 3, – 6

**Question 3.**

**a² x² – b² = 0**

**Solution:**

a² x² – b² = 0

⇒ (ax)² – (b)² =0

⇒ (ax + b) (ax – b) =

Either ax + b = 0, then x = \(\frac { -b }{ a }\)

or ax – b = 0. then x = \(\frac { b }{ a }\)

x = \(\frac { b }{ a }\) , \(\frac { -b }{ a }\)

**Question 4.**

**Solution:**

**Question 5.**

**x + \(\frac { 4 }{ x }\) = – 4; x ≠ 0**

**Solution:**

x + \(\frac { 4 }{ x }\) = -4

⇒ x² + 4 = -4x

⇒ x² + 4x + 4 = 0

⇒ (x + 2)² = 0

⇒ x + 2 = 0

⇒ x = – 2

**Question 6.**

**2x ^{4} – 5x² + 3 = 0**

**Solution:**

2x

^{4}– 5x² + 3 = 0

⇒ 2(x²)² – 5x² + 3 = 0

⇒ 2(x²)² – 3x² – 2x² + 3 = 0

⇒ 2x

^{4}– 3x² – 2x² + 3 = 0

⇒ x² (2x² – 3) – 1 (2x² – 3) = 0

⇒ (2x² – 3) (x² – 1) = 0

**Question 7.**

**x ^{4} – 2x² – 3 = 0**

**Solution:**

x

^{4}– 2x² – 3 = 0

⇒ (x²)² – 2x² – 3 = 0

⇒ (x²)² – 3x² + x² – 3 = 0

⇒ x² (x² – 3) + 1 (x² -3) = 0

⇒ (x² – 3) (x² + 1) = 0

Either x² – 3 = 0, then x² = 3 ⇒ x = √3

or x² + 1 = 0, then x² = – 1 In this case roots are not real

x = ±√3 or √3 , – √3

**Question 8.**

**Solution:**

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

**Question 11.**

**(x² + 5x + 4)(x² + 5x + 6) = 120**

**Solution:**

Let x² + 5x + 4 = y then x² + 5x + 6 = y + 2

Now (x² + 5x + 4) (x² + 5x + 6) = 120

⇒ y (y + 2) – 120 = 0

⇒ y² + 2y – 120 = 0

⇒ y² + 12y – 10y – 120 = 0

⇒ y (y + 12) – 10 (y + 12) = 0

⇒ (y + 12) (y – 10) = 0

Either y + 12 = 0, then y = – 12

or y – 10 = 0, then y = 10

(i) when y = -12, then x² + 5x + 4 = -12

⇒ x² + 5x + 4 + 12 = 0

⇒ x² + 5x + 16 = 0

Here a = 1, b = 5, c = 16

D = b² – 4ac = (5)² – 4 x 1 x 16 = 25 – 64 = -39

D < 0, then roots are not real

(ii) When y = 10, then x² + 5x + 4 = 10

⇒ x² + 5x + 4 – 10 = 0

⇒ x² + 5x – 6 = 0

⇒ x² + 6x – x – 6 = 0

⇒ x (x + 6) – 1 (x + 6) = 0

⇒ (x + 6) (x – 1) = 0

Either x + 6 = 0, then x = – 6

or x – 1 = 0, then x = 1

x = 1, -6

**Question 12.**

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

**(i) x² – 5x – 10 = 0 [2005]**

**(ii) 3x² – x – 7 = 0 [2004]**

**Solution:**

(i) Given Equation is : x² – 5x – 10 = 0

On comparing with, ax² + bx + c = 0

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

**Question 13.**

**Solution:**

**Question 14.**

**Solve:**

**(i) x² – 11x – 12 = 0; when x ∈ N**

**(ii) x² – 4x – 12 = 0; when x ∈ I**

**(iii) 2x² – 9x + 10 = 0; when x ∈ Q.**

**Solution:**

(i) x² – 11x – 12 = 0

⇒ x² – 12x + x – 12 = 0

⇒ x (x – 12) + 1 (x – 12) = 0

⇒ (x – 12) (x + 1) = 0

Either x – 12 = 0, then x = 12

or x + 1 = 0, then x = -1

x ∈ N

x = 12

(ii) x² – 4x – 12 = 0

⇒ x² – 6x + 2x – 12 = 0

⇒ x (x – 6) + 2 (x – 6)=0

⇒ (x – 6) (x + 2) = 0

Either x – 6 = 0, then x = 6

or x + 2 = 0, then x = -2

x ∈ I

x = 6, -2

(iii) 2x² – 9x + 10 = 0

⇒ 2x² – 4x – 5x + 10 = 0

⇒ 2x (x – 2) – 5 (x – 2) = 0

⇒ (x – 2) (2x – 5) = 0

Either x – 2 = 0, then x = 2

or 2x – 5 = 0, then 2x = 5 ⇒ x = \(\frac { 1 }{ 2 }\)

x ∈ Q

x = 2, \(\frac { 5 }{ 2 }\) or 2, 2.5

**Question 15.**

**Solve: (a + b)² x² – (a + b) x – 6 = 0, a + b ≠ 0.**

**Solution:**

(a + b)² x² – (a + b) x – 6 = 0

Let (a + b) x = y, then y² – y – 6 = 0

⇒ y² – 3y + 2y – 6 = 0

⇒ y (y – 3) + 2 (y – 3) = 0

⇒ (y – 3) (y + 2) = 0

Either y – 3 = 0, then y = 3

or y + 2 = 0, then y = – 2

(i) If y = 3, then

**Question 16.**

**Solution:**

Either x + p = 0, then x = -p

or x + q = 0, then x = -q

Hence x = -p, -q

**Question 17.**

**Solution:**

(i) x (x + 1) + (x + 2) (x + 3) = 42

⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0

⇒ 2x² + 6x – 36 = 0

⇒ x² + 3x – 18 = 0

⇒ x² + 6x – 3x – 18 = 0

⇒ x (x + 6) – 3(x + 6) = 0

⇒ (x + 6) (x – 3) = 0

Either x + 6 = 0, then x = -6

or x – 3 = 0, then x = 3

Hence x = 3, -6

**Question 18.**

**For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation:**

**(i) (m – 3) x² – 4x + 1 = 0**

**(ii) 3x² + 12x + (m + 7) = 0**

**(iii) x² – (m + 2) x + (m + 5) = 0**

**Solution:**

**Question 19.**

**Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.**

**Solution:**

px² – 4x + 3 = 0 …..(i)

Compare (i) with ax² + bx + c = 0

Here a = p, b = -4, c = 3

D = b² – 4ac = (-4)² – 4.p.(3) = 16 – 12p

As roots are equal, D = 0

16 – 12p = 0

⇒ \(\frac { 16 }{ 12 }\) = p

⇒ p = \(\frac { 4 }{ 3 }\)

**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:**

x² + 2 (m – 1) x + (m + 5) = 0.

Here, a = 1, b = 2 (m – 1), c = m + 5

So, discriminant, D = b² – 4ac

= 4(m – 1)² – 4 x 1 (m + 5)

= 4m² + 4 – 8m – 4m – 20

= 4m² – 12m – 16

For real and equal roots D = 0

So, 4m² – 12m – 16 = 0

⇒ m² – 3m – 4 = 0 (Dividingby4)

⇒ m² – 4m + m – 4 = 0

⇒ m (m – 4) + 1 (m – 4) = 0

⇒ (m – 4) (m + 1) = 0

⇒ m = 4 or m = -1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D are helpful to complete your math homework.

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