Question 382368
{{{sqrt(x^2 - 24) = sqrt(x + 6)}}}
We need to get rid of the square roots so we can "get at" the variable. So we start by squaring both sides of the equation:
{{{(sqrt(x^2 - 24))^2 = (sqrt(x + 6))^2}}}
This simplifies to:
{{{x^2 - 24 = x + 6}}}
This is now a quadratic equation. So we want one side of the equation to be zero. Subtracting x and 6 from each side we get:
{{{x^2-x-30 = 0}}}
Now we factor (or use the Quadratic Formula). This factors pretty easily:
(x-6)(x+5) = 0
From the Zero Product Property we know that this (or any) product can be zero <i>only</i> if one (or more) of the factors is zero. So
x-6 = 0 or x+5 = 0
Solving these we get:
x = 6 or x = -5<br>
Whenever you solve square root equations like yours, you square both sides of the equation. This is not wrong. But doing so can introduce what are called extraneous solutions. Extraneous solutions are solutions that work in the squared equation but do not work in the original equation. For this reason we <i>must</i> check our answers.<br>
When checking answers use the original equation:
{{{sqrt(x^2 - 24) = sqrt(x + 6)}}}
Checking x = 6:
{{{sqrt((6)^2 - 24) = sqrt((6) + 6)}}}
{{{sqrt(36 - 24) = sqrt(6 + 6)}}}
{{{sqrt(12) = sqrt(12)}}} Check!<br>
Checking x = -5:
{{{sqrt((-5)^2 - 24) = sqrt((-5) + 6)}}}
{{{sqrt(25-24) = sqrt(-5 + 6)}}}
{{{sqrt(1) = sqrt(1)}}} Check!<br>