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

Mar 14, 2016

See explanation...

#### Explanation:

I will use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 x + 5\right)$ and $b = 7$ later...

If I see an odd middle coefficient and am asked to "complete the square", I tend to groan slightly inwardly at the prospect of messy fractions.

Happily we can avoid some of the mess by multiplying our equation through by ${2}^{2} = 4$ first:

$0 = 4 \left({x}^{2} + 5 x - 6\right)$

$= 4 {x}^{2} + 20 x - 24$

$= {\left(2 x + 5\right)}^{2} - 25 - 24$

$= {\left(2 x + 5\right)}^{2} - 49$

$= {\left(2 x + 5\right)}^{2} - {7}^{2}$

$= \left(\left(2 x + 5\right) - 7\right) \left(\left(2 x + 5\right) + 7\right)$

$= \left(2 x - 2\right) \left(2 x + 12\right)$

$= 4 \left(x - 1\right) \left(x + 6\right)$

So the roots are: $x = 1$ and $x = - 6$