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:

$\sqrt{2} {x}^{2} + 7x +5 \sqrt{2} \\ \\ \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2} \\ \\ \sqrt{2} x(x + \sqrt{2} ) + 5(x + \sqrt{2}) \\ \\ (\sqrt{2} x + 5)(x + \sqrt{2} )$

$100 {x}^{2} - 20x + 1 \\ \\ 100 {x}^{2} - 10x - 10x + 1 \\ \\ 10x(10x - 1) - 1(10x - 1) \\ \\ (10x - 1)(10x - 1)$

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