Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(E)

Some equations may not look like a quadratic equation but can be reducible to quadratic equations. This reduction is done by proper substitution. Thus, in this exercise students will be solving different kinds of quadratic equations. For quick reference to any problem, the Selina Solutions for Class 10 Maths can be used. The solutions of this exercise are available in the Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations Exercise 5(E), PDF provided in the links below.

Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(E) Download PDF

 

selina solution concise maths class 10 chapter 5e
selina solution concise maths class 10 chapter 5e
selina solution concise maths class 10 chapter 5e
selina solution concise maths class 10 chapter 5e
selina solution concise maths class 10 chapter 5e

 

Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations

Exercise 5(A) Solutions

Exercise 5(B) Solutions

Exercise 5(C) Solutions

Exercise 5(D) Solutions

Exercise 5(F) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(E)

1. Solve each of the following equations:

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 1

Solution:

Given equation,

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 2

4x2 + 6x + x – 3 + 3x + 9 = 0

4x2 + 10x + 6 = 0

4x2 + 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 = ± 9

2x = ± 9 – 3

x = (± 9 – 3)/ 2

So,

x = (9 – 3)/ 2 or (-9 – 3)/ 2

Therefore,

x = 3 or x = -6

3. a2x2 – b2 = 0

Solution:

Given equation, a2x2 – b2 = 0

(ax)2 – b2 = 0

(ax + b)(ax – b) = 0

So,

ax + b = 0 or ax – b = 0

Therefore,

x = -b/a or b/a

4. x2 – 11/4 x + 15/8 = 0

Solution:

Given equation, x2 – 11/4 x + 15/8 = 0

Taking L.C.M we have,

(8x2 – 22x + 15)/ 8 = 0

8x2 – 22x + 15 = 0

8x2 – 12x – 10x + 15 = 0

4x(2x – 3) – 5(2x – 3) = 0

(4x – 5)(2x – 3) = 0

So, 4x – 5 = 0 or 2x – 3 = 0

Therefore,

x = 5/4 or x = 3/2

5. x + 4/x = -4; x ≠ 0

Solution:

Given equation, x + 4/x = -4

(x2 + 4)/ x = -4

x2 + 4 = -4x

x2 + 4x + 4 = 0

x2 + 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. 2x4 – 5x2 + 3 = 0

Solution:

Given equation, 2x4 – 5x2 + 3 = 0

Let’s take x2 = y

Then, the equation becomes

2y2 – 5y + 3 = 0

2y2 – 2y – 3y + 3 = 0

2y(y – 1) – 3(y – 1) = 0

(2y – 3) (y – 1) = 0

So, 2y – 3 = 0 or y – 1 = 0

y = 3/2 or y = 1

And, we have taken y = x2

Thus,

x2 = 3/2 or x2 = 1

x = ± √(3/2) or x = ±1

7. x4 – 2x2 – 3 = 0

Solution:

Given equation, x4 – 2x2 – 3 = 0

Let’s take x2 = y

Then, the equation becomes

y2 – 2y – 3 = 0

y2 – 3y + y – 3 = 0

y(y – 3) + 1(y – 3) = 0

(y + 1)(y – 3) = 0

So, y + 1 = 0 or y – 3 = 0

y = -1 or y = 3

And, we have taken y = x2

Thus,

x2 = – 1(impossible, no real solution)

x2 = 3

x = ± √3

8. Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 3

Solution:

Let us take (x + 1/x) = y …. (1)

Now, squaring it on both sides

(x + 1/x)2 = y2

x2 + 1/x2 + 2 = y2

So,

x2 + 1/x2 = y2 – 2 ….. (2)

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

9(y2 – 2) – 9(y) – 52 = 0

9y2 – 18 – 9y – 52 = 0

9y2 – 9y – 70 = 0

9y2 – 30y + 21y – 70 = 0

3y(3y – 10) + 7(3y – 10) = 0

(3y + 7)(3y – 10) = 0

So, 3y + 7 = 0 or 3y – 10 = 0

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

Now,

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

(x2 + 1)/x = -7/3 or (x2 + 1)/x = 10/3

3x2 – 10x + 3 = 0 or 3x2 + 7x + 3 = 0

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 43x2 – 9x – x + 3 = 0 or

3x(x – 3) – 1(x – 3) = 0

(3x – 1)(x – 3) = 0

So, x = 1/3 or 3

9. Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 5

Solution:

Let us take (x + 1/x) = y …. (1)

Now, squaring it on both sides

(x + 1/x)2 = y2

x2 + 1/x2 + 2 = y2

So,

x2 + 1/x2 = y2 – 2 ….. (2)

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

2(y2 – 2) – (y) = 11

2y2 – 4 – y = 11

2y2 – y – 15 = 0

2y2 – 6y + 5y – 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

(x2 + 1)/x = -5/2 or (x2 + 1)/x = 3

2(x2 + 1) = -5x or x2 + 1 = 3x

2x2 + 5x + 2 = 0 or x2 – 3x + 1 = 0

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 62x2 + 4x + x + 2 = 0 or

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

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

Hence, x = -1/2 or -2

10. Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 7

Solution:

Let us take (x – 1/x) = y …. (1)

Now, squaring it on both sides

(x – 1/x)2 = y2

x2 + 1/x2 – 2 = y2

So,

x2 + 1/x2 = y2 + 2 ….. (2)

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

(y2 + 2) – 3(y) – 2 = 0

y2 -3y = 0

y(y – 3) = 0

So, y = 0 or y – 3 = 0

Now,

(x – 1/x) = 0 or (x – 1/x) = 3

x2 – 1 = 0 or x2 – 1 = 3x

x2 = 1 or x2 – 3x – 1 = 0

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(E) - 8Therefore,

x = ± 1 or


Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class