Question 298755
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{{{sqrt(x+6)+sqrt(2-x)=4}}}

Pick one of the radicals and label it R.

If we pick the first radical to label R then we have
{{{R=sqrt(x+6)=R}}}.  So substituting R for the first
radical:

{{{R+sqrt(2-x)=4}}}

Isolate the radical term by subtracting R from both sides::

{{{sqrt(2-x)=4-R}}}

Square both sides:

{{{(sqrt(2-x))^2=(4-R)^2}}}

{{{2-x=(4-R)(4-R)}}}

FOIL out the right sides:

{{{2-x=16-4R-4R+R^2}}}

{{{2-x=16-8R+R^2}}}

Substitute {{{sqrt(x+6)}}} for R

{{{2-x=16-8sqrt(x+6)+(sqrt(x+6))^2}}}

{{{2-x=16-8sqrt(x+6)+x+6}}}

Isolate the radical term and simplify the above:


{{{8sqrt(x+6)=20+2x}}}

To make things easier divide every term by 2

{{{4sqrt(x+6)=10+x}}}

Square both sides:

{{{(4sqrt(x+6))^2=(10+x)^2}}}

{{{4^2*(sqrt(x+6))^2=(10+x)(10+x)}}}

{{{16(x+6)=100+10x+10x+x^2}}}

{{{16x+96=100+20x+x^2}}}

Get 0 on the left side:

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

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

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

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

{{{x=-2}}}

But we must ALWAYS check equations with even root
radicals as we often get extraneous solutions:

Checking:
{{{sqrt(x+6)+sqrt(2-x)=4}}}
{{{sqrt((-2)+6)+sqrt(2-(-2))=4}}}
{{{sqrt(4)+sqrt(2+2)=4}}}
{{{2+sqrt(4)=4}}}
{{{2+2=4}}}
{{{4=4}}}

It checks, so the solution is {{{x=-2}}} 

Edwin<</pre>