Question

${x}^{2} + x - 20 = 0$

Answer 1

$x^2 +x -20 = 0$

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

$x^2 + 5x -4x -20 = 0$

x(x+5) -4(x+5) = 0

(x+5)(x-4) = 0


x + 5 = 0   Or x -4 = 0 


x = -5  Or    x = 4    [ By Zero-Product Rule]

i hope this will help you


-by ABHAY

Answer 2

Answer:

______________________________

${x}^{2} + x - 20 = 0$

Term to be added:

$= ( \frac{1}{2} \times coefficient \: of \: x \: )^{2} \\ \\ = ( \frac{1}{2} \times 1)^{2} \\ \\ = (\frac{1}{2} )^{2} \\ \\ = \frac{1}{4}$

Therefore,

${x}^{2} + x + \frac{1}{4} - \frac{1}{4} - 20 = 0 \\ \\ (x + \frac{1}{2})^{2} = \frac{1}{4} + 20 \\ \\ (x + \frac{1}{2})^{2} = \frac{1 + 80}{4} \\ \\ (x + \frac{1}{2})^{2} = \frac{81}{4}$

$x+\frac{1}{2}=\pm\dfrac{9}{2}.....(taking \: of \: square \: roots) \\ \\ x + \frac{1}{2} = \frac{9}{2} or \: x + \frac{1}{2} = - \frac{9}{2} \\ \\ x = \frac{9}{2} - \frac{1}{2} \: or \: x = - \frac{9}{2} - \frac{1}{2} \\ \\ x = \frac{9 - 1}{2} \: \: \: or \: x = \frac{ - 9 - 1}{2} \\ \\ x = \frac{8}{2} \: or \: x = \frac{ - 10}{2} \\ \\ x = 4 \: \: or \: \: x = - 5$

Therefore,

The roots of the quadratic equation are 4 or -5.