Question 85463
Solve:
{{{sqrt(x+6) + sqrt(2-x) = 4}}} First, separate the square roots. Subtract {{{sqrt(2-x)}}} from both sides.
{{{sqrt(x+6) = 4-sqrt(2-x)}}} Next, square both sides.
{{{x+6 = 16-8*sqrt(2-x)+(2-x)}}} Simplify so that you have only the radical {{{sqrt(2-x)}}} (or a multiple thereof) on one side. 
{{{x+6 = 18-8*sqrt(2-x)-x}}} Add x to both sides.
{{{2x+6 = 18-8*sqrt(2-x)}}} Subtract 18 from both sides.
{{{2x-12 = -8*sqrt(2-x)}}} Now square both sides again.
{{{4x^2 - 48x + 144 = 64(2-x)}}} Simplify this so that you have a quadratic equation in standard form.
{{{4x^2 - 48x + 144 = 128-64x}}} Add 64x to both sides.
{{{4x^2 + 16x +144 = 128}}} Subtract 128 from both sides.
{{{4x^2 + 16x + 16 = 0}}} Factor out a 4 to simplify this.
{{{4(x^2 + 4x + 4) = 0}}} Now factor the parentheses.
{{{(x+2)(x+2) = 0}}} Applying the zero products principle, you get the double root:
{{{x = -2}}}

Check: 
{{{sqrt(x+6) + sqrt(2-x) = 4}}} Set x = -2.
{{{sqrt(-2+6) + sqrt(2-(-2)) = 4}}} Simplify.
{{{sqrt(4) + sqrt(4) = 4}}}
{{{2 + 2 = 4}}}
{{{4 = 4}}}