Answer
$x = 5$
Work Step by Step
Isolate the more complex radical:
$\sqrt {2x + 6} = \sqrt {x - 1} + 2$
Square both sides of the equation to eliminate the radical:
$\left(\sqrt {2x + 6}\right)^2 = \left(\sqrt {x - 1} + 2\right)^2\\
2x + 6 = (\sqrt {x - 1} + 2)^2$
Use the formula $(a+b)^2=a^2+2ab+b^2$ to expand the right side of the equation:
$2x + 6 = (x - 1) + 4\sqrt {x - 1} + 4$
$2x + 6 = x + 3 + 4\sqrt {x - 1}$
Move all terms except the radical to the left side of the equation:
$2x + 6 - x - 3 = 4\sqrt {x - 1}\\
x + 3 = 4\sqrt {x - 1}$
Square both sides of the equation to get rid of the radical:
$(x + 3)^2 = (4\sqrt {x - 1})^2$
$x^2 + 6x + 9 = (4)^2(\sqrt {x - 1})^2$
$x^2 + 6x + 9 = 16(x - 1)$
$x^2 + 6x + 9 = 16x - 16$
Move all terms to the left side of the equation:
$x^2 + 6x + 9 - 16x + 16=0\\
x^2 - 10x + 25 = 0$
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 $25$ and $b$ is $-10$. The factors $-5$ and $-5$ will work.
Let's rewrite the equation in factored form:
$(x - 5)(x - 5) = 0$
Solve the equation using the Zero-Product Property by equating each unique factor to $0$< then solve each equation:
$x - 5 = 0$
$x = 5$
To check if we have an extraneous solution, we substitute the solution into the original equation to see if the two sides equal one another.
Let's check the first solution:
$\sqrt {2(5) + 6} - \sqrt {5 - 1} = 2$
Combine the terms inside the radicand:
$\sqrt {16} - \sqrt {4} = 2$
Take the square roots:
$4 - 2 = 2$
Combine like terms:
$2 = 2$
Both sides are equal; therefore, this solution is valid.
