Question

root of the given quadratic equation x^2+x-20=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\red{\rm :\longmapsto\: {x}^{2} + x - 20 = 0}$

We use method of splitting of middle terms, to get the values of x.

$\red{\rm :\longmapsto\: {x}^{2} + 5x - 4x - 20 = 0}$

$\red{\rm :\longmapsto\:x(x + 5) - 4(x + 5) = 0}$

$\red{\rm :\longmapsto\:(x + 5)(x - 4) = 0}$

$\red{\bf\implies \: \boxed{ \:\bf{ \:x \: = \: - \: 5 \: \: \: or \: \: \: x \: = \: 4 \: }}}$

Alternative Method :-

Given quadratic equation is

$\green{\rm :\longmapsto\: {x}^{2} + x - 20 = 0}$

Using quadratic formula, we have

$\green{ \boxed{ \sf{ \:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \: }}}$

So, here

$\green{\rm :\longmapsto\:a = 1}$

$\green{\rm :\longmapsto\:b = 1}$

$\green{\rm :\longmapsto\:c = - \: 20}$

So, on substituting the values, we get

$\green{\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)( - 20) } }{2(1)}}$

$\green{\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{1 + 80} }{2}}$

$\green{\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{81} }{2}}$

$\green{\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: 9 }{2}}$

$\green{\rm :\longmapsto\:x = \dfrac{ - 1 \: + \: 9 }{2} \: \: \: or \: \: \: x = \dfrac{ - 1 \: - \: 9 }{2}}$

$\green{\rm :\longmapsto\:x = \dfrac{8}{2} \: \: \: or \: \: \: x = \dfrac{ - 10}{2}}$

$\green{\bf\implies \: \boxed{ \bf{ \:x \: = \: 4 \: \: \: or \: \: \: x \: = \: - \: 5 \: }}}$

Alternative Method :-

Method of Completing squares

Given quadratic equation is

$\rm :\longmapsto\: {x}^{2} + x - 20 = 0$

$\rm :\longmapsto\: {x}^{2} + x = 20$

Multiply by 4, on both sides, we get

$\rm :\longmapsto\: {4x}^{2} + 4x = 80$

On adding 1 on both sides, we get

$\rm :\longmapsto\: {4x}^{2} + 4x + 1 = 80 + 1$

$\rm :\longmapsto\: {(2x + 1)}^{2} = 81$

$\rm :\longmapsto\:2x + 1 = \pm \: 9$

$\rm :\longmapsto\:2x = - 1 \pm \: 9$

$\rm :\longmapsto\:2x = - 1 + 9 \: \: \: or \: \: \: 2x = -1 - 9$

$\rm :\longmapsto\:2x = 8 \: \: \: or \: \: \: 2x = -10$

$\purple{\bf\implies \: \boxed{ \bf{ \:x \: = \: 4 \: \: \: or \: \: \: x \: = \: - \: 5 \: }}}$