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

Not all quadratic equations can be solved easily by factorization method. That’s when solving quadratic equations using formula came into place. This method primarily uses the discriminant into the formula. The Selina Solutions for Class 10 Maths is the key tool for reference and doubt clearance for students. Students can also access the Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations Exercise 5(D) PDF from the links given below.

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

 

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

 

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(E) Solutions

Exercise 5(F) Solutions

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

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

(i) x2 – 6x = 27

Solution:

Given equation, x2 – 6x = 27

x2 – 6x – 27 = 0

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

By quadratic formula, we have

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

Therefore, x = 9 or -3

(ii) x2 – 10x + 21 = 0

Solution:

Given equation, x2 – 10x + 21 = 0

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

By quadratic formula, we have

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

Therefore, x = 7 or x = 3

(iii) x2 + 6x – 10 = 0

Solution:

Given equation, x2 + 6x – 10 = 0

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

By quadratic formula, we have

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

Therefore, x = -3 + √19 or x = -3 – √19

(iv) x2 + 2x – 6 = 0

Solution:

Given equation, x2 + 2x – 6 = 0

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

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 4

Therefore, x = -1 + √7 or x = -1 – √7

(v) 3x2 + 2x – 1 = 0

Solution:

Given equation, 3x2 + 2x – 1 = 0

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

By quadratic formula, we have

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

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

(vi) 2x2 + 7x + 5 = 0

Solution:

Given equation, 2x2 + 7x + 5 = 0

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

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 6

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

(vii) 2/3 x = -1/6 x2 – 1/3

Solution:

Given equation, 2/3 x = -1/6 x2 – 1/3

1/6 x2 + 2/3 x + 1/3 = 0

Multiplying by 6 on both sides

x2 + 4x + 2 = 0

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

By quadratic formula, we have

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

Therefore, x = -2 + √2 or x = -2 – √2

(viii) 1/15 x2 + 5/3 = 2/3 x

Solution:

Given equation, 1/15 x2 + 5/3 = 2/3 x

1/15 x2 – 2/3 x + 5/3 = 0

Multiplying by 15 on both sides

x2 – 10x + 25 = 0

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

By quadratic formula, we have

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

Therefore, x = 5 (equal roots)

(ix) x2 – 6 = 2 √2 x

Solution:

Given equation, x2 – 6 = 2 √2 x

x2 – 2√2 x – 6 = 0

Here, a = 1, b = -2√2 and c = -6

By quadratic formula, we have

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

Therefore, x = 3√2 or x = -√2

(x) 4/x – 3 = 5/ (2x + 3)

Solution:

Given equation, 4/x – 3 = 5/ (2x + 3)

(4 – 3x)/ x = 5/ (2x + 3)

On cross multiplying, we have

(4 – 3x)(2x + 3) = 5x

8x – 6x2 + 12 – 9x = 5x

6x2 + 6x – 12 = 0

Dividing by 6, we get

x2 + x – 2 = 0

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

By quadratic formula, we have

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

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)

2x2 + 4x + 3x + 6 = x2 + 3x + 4x + 12

2x2 + 7x + 6 = x2 + 7x + 12

x2 + 0x – 6 = 0

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

By quadratic formula, we have

Therefore, x = √6 or x = -√6

(xii) √6x2 – 4x – 2√6 = 0

Solution:

Given equation, √6x2 – 4x – 2√6 = 0

Here, a = √6, b = -4 and c = -2√6

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 12

Therefore, x = √6 or -√6/3

(xiii) 2x/ x – 4 + (2x – 5)/(x – 3) =Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 13

Solution:

Given equation, 2x/ x – 4 + (2x – 5)/(x – 3) =Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 14

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 15

25x2 – 175x + 300 = 12x2 – 57x + 60

13x2 – 118x + 240 = 0

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

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 16

Therefore, x = 6 or x = 40/13

(xiv)Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 17 

Solution:

From the given equation,

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 18

10x2 – 60x + 80 = 6x2 – 30x + 30

4x2 – 30x + 50 = 0

2x2 – 15x + 25 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 19

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) x2 – 8x +5 = 0

(ii) 5x2 + 10x – 3 = 0

Solution:

(i) x2 – 8x + 5 = 0

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

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 20

x = 4 ± 3.3

Thus, x = 7.7 or x = 0.7

(ii) 5x2 + 10x – 3 = 0

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

By quadratic formula, we have

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 21
Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 21

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) 2x2 – 10x + 5 = 0

Solution:

Given equation, 2x2 – 10x + 5 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 22

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

4x2 + 13x + 6 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 23

Therefore, x = -0.56 or x = -2.70

(iii) 4x2 – 5x – 3 = 0

Solution:

Given equation, 4x2 – 5x – 3 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 24

Therefore, x = 1.70 or x = -0.44

(iv) x2 – 3x – 9 = 0

Solution:

Given equation, x2 – 3x – 9 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 25

Therefore, x = 4.85 or x = -1.85

(v) x2 – 5x – 10 = 0

Solution:

Given equation, x2 – 5x – 10 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 26

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) 3x2 – 12x – 1 = 0

(ii) x2 – 16 x +6 = 0

(iii) 2x2 + 11x + 4 = 0

Solution:

(i) Given equation, 3x2 – 12x – 1 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 27

Therefore, x = 4.082 or x = -0.082

(ii) Given equation, x2 – 16 x + 6 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 28

Therefore, x = 15.616 or x = 0.384

(iii) Given equation, 2x2 + 11x + 4 = 0

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

Concise Selina Solutions Class 10 Maths Chapter 5 ex. 5(D) - 29

Therefore, x = -0.392 or x = -5.110

5. Solve:

(i) x4 – 2x2 – 3 = 0

Solution:

Given equation, x4 – 2x2 – 3 = 0

x4 – 3x2 + x2 – 3 = 0

x2(x2 – 3) + 1(x2 – 3) = 0

(x2 + 1) (x2 – 3) = 0

So, x2 + 1 = 0 (which is not possible) or x2 – 3 = 0

Hence,

x2 – 3 = 0

x = ± √3

(ii) x4 – 10x2 + 9 = 0

Solution:

Given equation, x4 – 10x2 + 9 = 0

x4 – x2 – 9x2+ 9 = 0

x2(x2 – 1) – 9(x2 – 1) = 0

(x2 – 9)(x2 – 1) = 0

So, we have

x2 – 9 = 0 or x2 – 1 = 0

Hence,

x = ± 3 or x = ± 1


1 Comment

  1. Hiii nice …thnx a lot for it

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