Question 415764
Here's a procedure for solving these kinds of equations:<ol><li>Isolate a square that has the variable in its radicand. ("Radicand" is the name fo the expression within a radical.) Coefficients in front of the isolated square root are OK.</li><li>Square both sides of the equation.</li><li>If there is still a square root with a variable in its radicand, repeat steps 1 and 2.</li><li>At this point you should have an equation that has no square roots with the variable in its radciand. Use appropriate techniques to solve this equation.</li><li>Check you answer(s). <i>This is not optional!</i> Whenever you square both sides of an equation, like we have done a least once at step 2, extraneous solutions can be introduced. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. Extraneous solutions can happen <i>even if no mistakes were made.</i> So even expert mathematicians must check their answers on thse problems. Extraneous solutions, if any, must be rejected.</li></ol>
Let's see how this works on your equation:
{{{sqrt(x+7) = x-5}}}
1) Isolate a square root. 
Your only square root is already isolated,<br>
2) Square both sides:
{{{(sqrt(x+7))^2 = (x-5)^2}}}
The left side is easy to square. Since <i>exponents do not distribute</i> squaring the right side is a little harder than it may seem. To square the right side correctly we can use FOIL on (x-7)(x-7) or use the {{{(a-b)^2 = a^2-2ab+b^2}}} pattern with the "a" being x and the "b" being 5. I like to use the patterns:
{{{x+7 = (x)^2 -2(x)(5) + (5)^2}}}
which simplifies to:
{{{x+7 = x^2 -10x + 25}}}<br>
3) If there are square roots left...
There are no square roots left so we can proceed to step 4.<br>
4) Solve the equation.
The equation is a quadratic equation. So we want one side to be zero. Subtracting x and 7 from each side we get:
{{{0 = x^2 -11x + 18}}}
Now we factor (or use the Quadratic Formula). This factors easily:
0 = (x-9)(x-2)
From the Zero Product Property we know that one of the factors must be zero. So:
x-9 = 0 or x-2 = 0
Solving these we get:
x = 9 or x = 2<br>
5) Check your answer(s).
Always use the original equation to check:
{{{sqrt(x+7) = x-5}}}
Checking x = 9:
{{{sqrt((9)+7) = (9)-5}}}
which simplifies as follows:
{{{sqrt(16) = 4}}}
4 = 4 Check!<br>
Checking x = 2:
{{{sqrt((2)+7) = (2)-5}}}
which simplifies as follows:
{{{sqrt(9) = -3}}}
3 = -3 Check failed!<br>
So x = 2 is an extraneous solution which we must reject. (Again, this does not mean we made a mistake earlier.) So the only solution to your equation is x = 9.