# How do you solve sqrt(x^2 + 4) -sqrt(x + 10) = 0?

Aug 5, 2015

$\left\{\begin{matrix}x = - 2 \\ x = 3\end{matrix}\right.$

#### Explanation:

Notice that you are dealing with the difference of two terms, which can only be equal to zero if the two terms in question are equal to each other.

This means that you can write

$\sqrt{{x}^{2} + 4} - \sqrt{x + 10} = 0 \iff \sqrt{{x}^{2} + 4} = \sqrt{x + 10}$

Your possible solution(s) must satisfy the condition $x + 10 \ge 0$, since you cannot take the square root of a negative number if you're interested in a solution that's a real number.

So, you need

$x + 10 \ge 0 \implies x \ge - 10$

Square both sides of the equation to get rid of the radicals

${\left(\sqrt{{x}^{2} + 4}\right)}^{2} = {\left(\sqrt{x + 10}\right)}^{2}$

${x}^{2} + 4 = x + 10$

Move all the terms to one side of the equation and use the quadratic formula to determine the two solutions

${x}^{2} - x - 6 = 0$

${x}_{1 , 2} = \frac{\left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \cdot \left(- 6\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2} = \left\{\begin{matrix}{x}_{1} = \frac{1 + 5}{2} = 3 \\ {x}_{2} = \frac{1 - 5}{2} = - 2\end{matrix}\right.$

Since both values satisfy the condition $x \ge - 10$, they will be both valid solutions to the original equation.