Question -
Answer -
(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
Solutions:
(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 = 0or 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 = 0or 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