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

Solve each of the following equations :
Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.2

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.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.2

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.
2x4 – 5x² + 3 = 0
Solution:
2x4 – 5x² + 3 = 0
⇒ 2(x²)² – 5x² + 3 = 0
⇒ 2(x²)² – 3x² – 2x² + 3 = 0
⇒ 2x4 – 3x² – 2x² + 3 = 0
⇒ x² (2x² – 3) – 1 (2x² – 3) = 0
⇒ (2x² – 3) (x² – 1) = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q6.1

Question 7.
x4 – 2x² – 3 = 0
Solution:
x4 – 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.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.3

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.4

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.3

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.3

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.3

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q15.1

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.1
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.3
Either x + p = 0, then x = -p
or x + q = 0, then x = -q
Hence x = -p, -q

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.1
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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.2

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:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.3

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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