Answer
$x = 2$
Work Step by Step
Isolate the radical:
$\sqrt {x + 7} = x + 1$
Square both sides of the equation to eliminate the radical:
$x + 7 = (x + 1)^2$
$x + 7 = x^2 + 2x + 1$
Move all terms to the left side of the equation:
$x+7-x^2-2x-1=0\\
-x^2 - x + 6 = 0$
Divide both sides of the equation by $-1$ so that the $x^2$ term is positive:
$x^2 + x - 6 = 0$
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 $-6$ and $b$ is $1$. The factors $3$ and $-2$ will work.
Let's rewrite the equation in factored form:
$(x + 3)(x - 2) = 0$
Solve using the Zero-Product Property by equating each factor to $0$, then solve each equatoin:
First factor:
$x + 3= 0$
$x = -3$
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:
$\sqrt {-3 + 7} - (-3) = 1$
Combine the terms inside the radicand:
$\sqrt {4} - (-3) = 1$
Take the square root:
$2 + 3 = 1$
Combine like terms:
$5 = 1$
We can see that the two sides are not equal; therefore, this is an extraneous solution.
Let's try the solution $x = 2$:
$\sqrt {2 + 7} - 2 = 1$
Combine the terms inside the radicand:
$\sqrt {9} - 2 = 1$
Take the square root:
$3 - 2 = 1$
Combine like terms:
$1 = 1$
Both sides are equal; therefore, this solution is valid.
