```Question 345182
{{{sqrt(x-5)-sqrt(x-8)=3}}}
Here's a procedure for solving equations like this:<ol><li>Isolate a square root (i.e. get it by itself on one side of the equation).</li><li>Square both sides of the equation.</li><li>If there are still any square roots, repeat steps #1 and #2 until they are all gone.</li><li>At this point you have an equation without square roots. Use an appropriate procedure for solving the type of equation you now have.</li><li>Check your answer(s)!. This is important, not just a good idea, whenever you square both sides of an equation (like you have done at least once at step #2). Squaring both sides of an equation can introduce what are called extraneous solutions. These are solutions which work in the squared equation <i>but not in the original equation!</i> You must check your answers, using the original equation, and reject any that do not work.</li></ol>
Let's use this on yhour equation.
1) Isolate a square root. Add the second square root to both sides:
{{{sqrt(x-5) = sqrt(x-8) + 3}}}
2) Square both sides:
{{{(sqrt(x-5))^2 = (sqrt(x-8) + 3)^2}}}
Squaring the left side is simple. But squaring the right side can easily be done incorrectly. Exponent do not distribute! We must use FOIL or the {{{(a+b)^2 = a^2 + 2ab + b^2}}} pattern to square this correctly:
{{{x-5 = (sqrt(x-8))^2 + 2sqrt(x-8)(3) + 3^2}}}
{{{x-5 = x-8 + 6sqrt(x-8) + 9}}}
{{{x-5 = x + 6sqrt(x-8) + 1}}}
3) We still have a square root. So we need to repeat steps #1 and #2. Isolate a suqare root. There's only one left so we'll subtract x and 1 from each side to isolate it:
{{{-6 = 6sqrt(x-6)}}}
Square both sides:
{{{36 = 36(x-6)}}}
The square roots are gone so we can move on to step 4.
4) Solve the new equation. This is a simple equation to solve. We can start by dividing by 36:
{{{1 = x-6}}}
Add 6 to each side:
{{{7 = x}}}
5)> Check the answer(s) in the original equation.
{{{sqrt(x-5)-sqrt(x-8)=3}}}
Checking x = 7:
{{{sqrt((7)-5) - sqrt((7)-8) = 3}}}
{{{sqrt(2) - sqrt(-1) = 3}}}
At this point we should notice a problem. We cannot have a negative radicand in a square root! So x = 7 does NOT work in the original equation. It is an extraneous solution which we must reject.<br>
Since x = 7 was the only answer we found and since we had to reject it, there are no solutions to the original equation.```