Question

1. Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x +1/8 = 0
(v) 100x2 – 20x + 1 = 0

Answer 1

Answer:

(i)=>(x – 5)(x + 2)

(ii)=> x = -2 or x = 3/2

(iii)=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

(iv)⇒ x = 1/4 or x = 1/4

(v) (10x – 1) = 0 or (10x – 1) = 0

Step-by-step explanation:

(i) Given, x2 – 3x – 10 =0

Taking LHS,

=>x2 – 5x + 2x – 10

=>x(x – 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

=> x = 5 or x = -2

(ii) Given, 2x2 + x – 6 = 0

Taking LHS,

=> 2x2 + 4x – 3x – 6

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

(iii) √2 x2 + 7x + 5√2=0

Taking LHS,

=> √2 x2 + 5x + 2x + 5√2

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

(iv) 2x2 – x +1/8 = 0

Taking LHS,

=1/8 (16x2  – 8x + 1)

= 1/8 (16x2  – 4x -4x + 1)

= 1/8 (4x(4x  – 1) -1(4x – 1)

= 1/8 (4x – 1)2

The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0

Therefore, (4x – 1) = 0 or (4x – 1) =0

⇒ x = 1/4 or x = 1/4

(v) Given, 100x2 – 20x + 1=0

Taking LHS,

= 100x2 – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)2

The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0

∴ (10x – 1) = 0 or (10x – 1) = 0

want more:

1. Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3)

(ii) x2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x2 + 3x + 1 = (x – 2)2

(vii) (x + 2)3 = 2x (x2 – 1)

(viii) x3 – 4x2 – x + 1 = (x – 2)3

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Answer 2

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(i) Given, x2 – 3x – 10 =0

Taking LHS,

=>x2 – 5x + 2x – 10

=>x(x – 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

=> x = 5 or x = -2

(ii) Given, 2x2 + x – 6 = 0

Taking LHS,

=> 2x2 + 4x – 3x – 6

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

(iii) √2 x2 + 7x + 5√2=0

Taking LHS,

=> √2 x2 + 5x + 2x + 5√2

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

(iv) 2x2 – x +1/8 = 0

Taking LHS,

=1/8 (16x2 – 8x + 1)

= 1/8 (16x2 – 4x -4x + 1)

= 1/8 (4x(4x – 1) -1(4x – 1))

= 1/8 (4x – 1)2

The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

(v) Given, 100x2 – 20x + 1=0

Taking LHS,

= 100x2 – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)2

The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10