# How do you solve x^2+5x-6=0 by completing the square?

Jul 26, 2018

$x = 1 \mathmr{and} x = - 6$

#### Explanation:

Here,
${x}^{2} + 5 x - 6 = 0$
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To complete perfect square let $M$ be the third term,
such that ${x}^{2} + 5 x + M$ is a perfect square.

So,

color(blue)((i)1^(st) term=x^2
color(blue)((ii)2^(nd)term=5x
color(blue)((iii)3^(rd)term=M

Formula :
color(blue)(3^(rd)term=(2^(nd)term)^2/(4xx 1^(st) term)

$M = {\left(5 x\right)}^{2} / \left(4 \times {x}^{2}\right) = \frac{25 {x}^{2}}{4 {x}^{2}} = \frac{25}{4}$

$i . e . {x}^{2} + 5 x + \left(\frac{25}{4}\right)$ is a perfect square.
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So,

${x}^{2} + 5 x = 6$

=>x^2+5x+color(red)(25/4)=6+color(red)(25/4

$\implies {\left(x + \frac{5}{2}\right)}^{2} = \frac{49}{4} = {\left(\frac{7}{2}\right)}^{2}$

$\implies x + \frac{5}{2} = \pm \frac{7}{2}$

$\implies x + \frac{5}{2} = \frac{7}{2} \mathmr{and} x + \frac{5}{2} = - \frac{7}{2}$

$\implies x = \frac{7}{2} - \frac{5}{2} \mathmr{and} x = - \frac{7}{2} - \frac{5}{2}$

$\implies x = \frac{2}{2} \mathmr{and} x = - \frac{12}{2}$

$\implies x = 1 \mathmr{and} x = - 6$