Question

Solve the following eq
(1) x² + x - 20=0
​

Answer 1

$x^{2} + x - 20 = 0\\x^{2} + (5x - 4x) - 20 = 0\\\\x^{2} + 5x - 4x - 20 = 0\\\\x(x+5) - 4(x+5) = 0\\\\(x+5) (x-4) = 0\\\\x = -5\\x = 4$

Roots of this equation are -5 and 4

Answer 2

x²+x-20=0

comparing the above equation with ax²+bx+c=0

therefore, a=1, b=1 and c=-20

third term=(b/2) ²

=(1/2) ²

=1/4

therefore by completing square method,

x²+x+1/4-1/4-20=0

(x+1/2) ²-1/4-20=0

(x+1/2) ²-1-80/4=0

(x+1/2) ²-81/4=0

(x+1/2) ²=81/4

x±1/2=9/2

x=9/2+1/2 or x=9/2-1/2

x=-5 or 4

Therefore,the roots of the given quadratic equation is -5 or 4

Hope it will help you