Answer
$x = 9$
Work Step by Step
Isolate the radical:
$\sqrt {x + 7} = x - 5$
Square both sides of the equation to eliminate the radical:
$x + 7 = (x - 5)^2$
$x + 7 = x^2 - 10x + 25$
Move all terms to the left side of the equation:
$x+7-x^2+10x-25=0\\
-x^2 + 11x - 18 = 0$
Divide both sides of the equation by $-1$ so that the $x^2$ term is positive:
$x^2 - 11x + 18 = 0$
We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $ac$ is $18$ and $b$ is $-11$. The factors $-9$ and $-2$ will work.
Let's rewrite the equation in factored form:
$(x - 9)(x - 2) = 0$
Solve using the Zero-Product Property by equating eah factor to $0$, then solve each equation:
First factor:
$x - 9= 0$
$x = 9$
Second factor:
$x - 2 = 0$
$x = 2$
To check if we have any extraneous solutions, we substitute each solution into the original equation to see if the two sides equal one another.
Let's check the first solution:
$\sqrt {9 + 7} + 5 = 9$
Combine the terms inside the radicand:
$\sqrt {16} + 5 = 9$
Take the square root:
$4 + 5 = 9$
Combine like terms:
$9 = 9$
We can see that the two sides are equal; therefore, this is a valid solution.
Let's try the solution $x = 2$:
$\sqrt {2 + 7} + 5 = 2$
Combine the terms inside the radicand:
$\sqrt {9} + 5 = 2$
Take the square root:
$3 + 5 = 2$
Combine like terms:
$8 = 2$
Both sides are not equal; therefore, this solution is extraneous.
