Answer
$x = 6$
Work Step by Step
To get rid of radicals, we square both sides of the equation:
$(\sqrt {3x + 7})^2 = (x - 1)^2$
$3x + 7 = (x - 1)(x - 1)$
Use the FOIL method to expand the right side of the equation. With the FOIL method, we multiply the first terms first, then the outer terms, then the inner terms, and, finally, the last terms:
$3x + 7 = (x)(x) -1x - 1x + (-1)(-1)$
$3x + 7 = x^2 - 2x + 1$
Rewrite the equation so that one side is $0$:
$0=x^2 - 2x + 1 - 3x - 7\\
0=x^2-5x-6\\
x^2-5x-6=0$
Factor the trinomial:
$(x-6)(x+1)=0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation:
First factor:
$x - 6 = 0$
$x = 6$
Second factor:
$x + 1 = 0$
$x = -1$
To check if our solutions are correct, we plug our solutions back into the original equation to see if the left and right sides equal one another.
Let's plug in $x = 6$ first:
$(6)^2 - (5)(6) - 6 = 0$
Multiply first, according to order of operations:
$36 - 30 - 6 = 0$
Now add and subtract from left to right:
$6 - 6 = 0$
Subtract once again:
$0 = 0$
The left and right sides are equal; therefore, this solution is correct.
Let's check $x = -1$:
$(-1)^2 + 5(-1) - 6 = 0$
Multiply first, according to order of operations:
$1 - 5 - 6 = 0$
Now add and subtract from left to right:
$-4 - 6 = 0$
Subtract once again:
$-10 = 0$
The left and right sides are not equal; therefore, this is an extraneous solution.
The solution is $x = 6$.
