How do you complete the square for $x^2 + 12x$ ?

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Answer 1 · 2015-05-17T21:38:21

The answer is $x=0$ ; $x=-12$ .

Problem: Complete the square for $x^2+12x$ .

Rewrite the equation as a trinomial.

$x^2+12x+0$

Move 0 to the right-hand side.

$x^2+12x=0$

Divide the coefficient of the $x$ -term by 2, then square the result. Add the result to both sides.

$12/2=6$ ; $6^2=36$

$x^2+12x+36=36$

Factor the perfect square trinomial $x^2+12x+36$ on the left-hand side.

$(x+6)^2=36$

Take the square root of both sides and solve for $x$ .

$sqrt((x+6)^2)=+-sqrt36$ =

$x+6=+-6$

$x=-6+6=0$

$x=-6-6=-12$

Check

If $x=0$ :

$(0)^2+12(0)=0$

$0=0$

If $x=-12$ :

$(-12)^2+12(-12)=0$

$144-144=0$

$0=0$

How do you complete the square for x^2 + 12xx^2 + 12x?