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

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Answer 1 · 2016-03-14T20:50:58

See explanation...

I will use the difference of squares identity:

$a^2-b^2=(a-b)(a+b)$

with $a=(2x+5)$ 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:

$0 = 4(x^2+5x-6)$

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

$=(2x+5)^2-25-24$

$=(2x+5)^2-49$

$=(2x+5)^2-7^2$

$=((2x+5)-7)((2x+5)+7)$

$=(2x-2)(2x+12)$

$=4(x-1)(x+6)$

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

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