Question 354730
{{{(x^2+4x-4)^2=64}}}
<pre>
We use the principle of square roots:

Take away the square on the right, take ± square roots of the right side

{{{x^2+4x-4="" +- sqrt(64)}}}

{{{x^2+4x-4="" +- 8}}}

Make two equations, one with the + and one with the -

{{{x^2+4x-4=""+8}}} and {{{x^2+4x-4=-8}}}

Solving the first one

{{{x^2+4x-4=""+8}}}

Get 0 on the right:

{{{x^2+4x-12=0}}}

Factor the left side by thinking of two whole numbers which
have product 12 and difference 4.  They are 6 and 2.  So
we write

{{{"(x"}}} {{{"6)(x"}}} {{{"2)"=0}}}

Then we put in the signs so that if we were to FOIL it out
we would get the middle term {{{""+4x}}} in the middle.

{{{(x+6)(x-2)=0}}}

Then we use the zero-factor principle:

set each factor = 0

{{{x+6=0}}} gives solution {{{x=-6}}}

{{{x-2=0}}} gives solution {{{x=2}}}

-------------------

Solving the second one

{{{x^2+4x-4=-8}}}

Get 0 on the right:

{{{x^2+4x+4=0}}}

Factor the left side by thinking of two whole numbers which
have product 4 and sum 4.  They are 2 and 2.  So
we write

{{{"(x"}}} {{{"2)(x"}}} {{{"2)"=0}}}

Then we put in the signs so that if we were to FOIL it out
we would get the middle term {{{""+4x}}} in the middle.

{{{(x+2)(x+2)=0}}}

Then we use the zero-factor principle:

set each factor = 0

{{{x+2=0}}} gives solution {{{x=-2}}}

The second factor is the same, so we do not get an additional solution.

Checking in the original equation:

Checking {{{x=-6}}}

{{{(x^2+4x-4)^2=64}}}

{{{((-6)^2+4(-6)-4)^2=64}}}
{{{(36-24-4)^2=64}}}
{{{(8)^2=64}}}
{{{64=64}}}

Checking {{{x=2}}}

{{{(x^2+4x-4)^2=64}}}

{{{((2)^2+4(2)-4)^2=64}}}
{{{(4+8-4)^2=64}}}
{{{(8)^2=64}}}
{{{64=64}}}

Checking {{{x=-2}}}

{{{(x^2+4x-4)^2=64}}}

{{{((-2)^2+4(-2)-4)^2=64}}}
{{{(4-8-4)^2=64}}}
{{{(-8)^2=64}}}
{{{64=64}}}

They are all solutions.

Edwin</pre>