Answer
$x = 8$
Work Step by Step
First, we want to isolate the more complex radical:
$\sqrt {3x + 1} = \sqrt {x + 1} + 2$
Square both sides of the equation to eliminate the radical:
$3x + 1 = (\sqrt {x + 1} + 2)^2$
Use the formula $(a+b)^2=a^2+2ab+b^2$ to expand the right side of the equation:
$3x + 1 = (x + 1) + 4\sqrt {x + 1} + 4$
$3x + 1 = x + 5 + 4\sqrt {x + 1}$
Move all terms except the radical to the left side of the equation:
$3x + 1 - x - 5 =4\sqrt {x + 1}\\
2x - 4 = 4\sqrt {x + 1}$
Divide both sides by $2$ to simplify:
$x - 2 = 2\sqrt {x + 1}$
Square both sides of the equation to get rid of the radical:
$(x - 2)^2 = (2\sqrt {x + 1})^2$
$(x - 2)^2 = (2)^2(\sqrt {x + 1})^2$
$(x - 2)^2 = 4(x + 1)$
$(x - 2)^2 = 4x + 4$
Use the formula $(a-b)^2=a^2-2ab+b^2$ to expand the left side of the equation:
$x^2 - 4x + 4 = 4x+4$
Move all terms to the left side of the equation:
$x^2 - 4x + 4 - 4x-4=0\\
x^2 - 8x = 0$
Factor out any common terms:
$x(x - 8) = 0$
Solve using the Zero-Product Property by equating each factor to $0$, then solve each equation:
First factor:
$x = 0$
Second factor:
$x - 8 = 0$
$x = 8$
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:
$\sqrt {3(0) + 1} - \sqrt {0 + 1} = 2$
Combine the terms inside the radicand:
$\sqrt {4} - \sqrt {1} = 2$
Take the square roots:
$2 - 1 = 2$
Combine like terms:
$1 = 2$
We can see that the two sides are not equal; therefore, this is an extraneous solution.
Let's try the solution $x = 8$:
$\sqrt {3(8) + 1} - \sqrt {8 + 1} = 2$
Combine the terms inside the radicand:
$\sqrt {25} - \sqrt {9} = 2$
Take the square roots:
$5 - 3 = 2$
Combine like terms:
$2 = 2$
Both sides are equal; therefore, this solution is valid.
