Answer
$x = 2$
Work Step by Step
We don't want to work with radicals, so we need to transform them. To get rid of radicals, we square both sides of the equation:
$(\sqrt {x + 2})^2 = (x)^2$
Squaring the radical will leave just the binomial:
$x + 2 = x^2$
Rewrite the equation so the term with the greatest degree lies on the left side of the equation:
$x^2 = x + 2$
We move all terms to the left side of the equation so that we are left with $0$ on the right side of the equation:
$x^2 - x - 2 = 0$
We solve by factoring. To factor a quadratic equation in the form $ax^2 + bx + c = 0$, we look at factors of the product of $a$ and $c$ such that, when added together, equal $b$.
For the equation $x^2 - x - 2 = 0$, $(a)(c)$ is $(1)(-2)$, or $-2$, but when added together will equal $b$ or $-1$. One of the factors needs to be positive and the other one negative, but the negative factor should be the one with the greater absolute value. This is because a negative number multiplied with a positive number will equal a negative number; however, when a negative number is added to a positive number, the result can be either negative or positive, depending on which number has the greater absolute value.
We came up with the following possibility:
$(a)(c)$ = $(-2)(1)$
$b = -1$
Let's take a look at our factorization:
$(x - 2)(x + 1) = 0$
The zero product property states that if the product of two factors equals zero, then either one of the factors is zero or both factors equal zero. We can, therefore, set each factor to $0$:
First factor:
$x - 2 = 0$
Add $2$ to each side to solve for $x$:
$x = 2$
Second factor:
$x + 1 = 0$
Subtract $1$ from each side to solve for $x$:
$x = -1$
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:
$\sqrt {2 + 2} = 2$
Add what's inside the brackets:
$\sqrt {4} = 2$
Take the square root of $4$:
$2 = 2$
The left and right sides are equal; therefore, this solution is correct.
Let's check $x = -1$:
$\sqrt {-1 + 2} = -1$
Add numbers inside brackets first:
$\sqrt {1} = -1$
Take the square root of $1$:
$1 \ne -1$
The left and right sides are not equal; therefore, this is an extraneous solution.
The only solution is $x = 2$.
