Answer
$x = 2$ or $x = -1$
Work Step by Step
Isolate the more complex radical:
$\sqrt {3 - x} = -\sqrt {x + 2} + 3$
Square both sides of the equation to eliminate the radical:
$(\sqrt {3 - x})^2 = (-\sqrt {x + 2} + 3)^2\\
3 - x = (-\sqrt {x + 2} + 3)^2$
Use the FOIL method to distribute the right side of the equation:
$3 - x = (x + 2) - 6\sqrt {x + 2} + 9$
$3 - x = x + 11 - 6\sqrt {x + 2}$
Move all terms except the radical to the left side of the equation:
$3 - x - x - 11 = 6\sqrt {x + 2}\\
-8 - 2x = -6\sqrt {x + 2}\\
- 2x-8 = -6\sqrt {x + 2}$
Multiply all terms by $-1$:
$2x + 8 = 6\sqrt {x + 2}$
Square both sides of the equation to get rid of the radical:
$(2x + 8)^2 = (6\sqrt {x + 2})^2$
$(2x + 8)^2 = (6)^2(\sqrt {x + 2})^2$
$(2x + 8)^2 = 36(x + 2)$
$(2x + 8)^2 = 36x + 72$
Use the FOIL method to distribute on the left side of the equation:
$4x^2 + 32x + 64 = 36x+72$
Divide both sides by $4$:
$x^2 + 8x + 16 = 9x + 18$
Move all terms to the left side of the equation:
$x^2 + 8x + 16 - 9x - 18=0\\
x^2 - x - 2 = 0$
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 $-2$ and $b$ is $-1$. The factors $-2$ and $1$ will work.
Let's rewrite the equation in factored form:
$(x - 2)(x + 1) = 0$
Solve the equation using the Zero-Product Property by equating each factor to $0$, then solve each equation:
First factor:
$x - 2 = 0$
$x = 2$
Second factor:
$x + 1 = 0$
$x = -1$
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:
$\sqrt {3 - 2} + \sqrt {2 + 2} = 3$
Combine the terms inside the radicand:
$\sqrt {1} + \sqrt {4} = 3$
Take the square roots:
$1 + 2 = 3$
Combine like terms:
$3 = 3$
Both sides are equal; therefore, this solution is valid.
Let's check the second solution:
$\sqrt {3 - (-1)} + \sqrt {-1 + 2} = 3$
Combine the terms inside the radicand:
$\sqrt {4} + \sqrt {1} = 3$
Take the square roots:
$2 + 1 = 3$
Combine like terms:
$3 = 3$
Both sides are equal; therefore, this solution is valid.
