Question

Find the roots of quadratic equations by factorisation:

(i) √2 x2 + 7x + 5√2=0

(ii) 100x2 – 20x + 1 = 0​

Answer 1

(i) √2 x2 + 7x + 5√2=0

Considering the L.H.S. first,

⇒ √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 (√2x + 5)(x + √2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

⇒ x = -5/√2 or x = -√2

(ii) Given, 100x2 – 20x + 1=0

Considering the L.H.S. first,

⇒ 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

Therefore,

(10x – 1) = 0

or (10x – 1) = 0

⇒ x =1/10 or x =1/10

Answer 2

ANSWER:-

Solve:-

Given:-

(i)  √2x^2 + 7x + 5√2=0

Considering the L.H.S. first,

$\mathtt {\implies \sqrt{2x {}^{2} } + 5x + 2x + 5\sqrt{2} }$

$\mathtt{ \implies x( \sqrt{2x} + 5) + \sqrt{2} (\sqrt{2x} + 5)}$

$\mathtt{ = ( \sqrt{2x} + 5 )(x + \sqrt{2} )}$

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

Therefore:-

$\mathtt {\sqrt{2x} + 5 = 0 \: or \: x + \sqrt{2} = 0}$

$\mathtt\red {\implies x = - \frac{ - 5}{ \sqrt{2} } \: or \: x = - \sqrt{2} }$

Given:-

ii) 100x^2 – 20x + 1 = 0

Considering the L.H.S. first

$\mathtt{\implies 100x {}^{2} - 10x - 10x + 1}$

$\mathtt{ \implies 10x(10x – 1) -1(10x – 1)}$

$\mathtt{ = (10x - 1) {}^{2} }$

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

Therefore:-

(10x – 1) = 0

or (10x – 1) = 0

⇒ x =1/10 or x =1/10