How do you solve using the completing the square method $x^2 - 6x =13$ ?

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Answer 1 · 2016-05-06T05:11:01

$x=3+sqrt22$
$x=3-sqrt22$

Given -

$x^2-6x=13$
Divide the coefficient of $x$ by 2 and add it to both sides

$x^2-6x+((-6)/2)^2=13 ++((-6)/2)^2$

You have a perfect square on the left hand side

$x^2-6x+9=13+9$

Rewrite it as -

$(x-3)^2=22$

Taking square on both sides, we have

$(x-3)=+-sqrt22$

$x=3+sqrt22$
$x=3-sqrt22$

How do you solve using the completing the square method x^2 - 6x =13x^2 - 6x =13?