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

Jun 1, 2016

The only solution is $x = 6$.
See explanation below about arithmetic square root.

#### Explanation:

First of all, we assume that the equation should be solved for real numbers $x$. Any other choice for its argument $x$ (complex, integer, etc.) is usually specified explicitly.

When square root is involved and we plan to raise both sides of an equation to the power of 2 to get rid of it, we have to start from the domain where the equation can have sense. In this case, the set of real numbers where the solution should exist is defined as $x \ge - 10$ for the square root to be defined.

Another restriction is related to the fact that, when we use a notation $\sqrt{A}$, we assume that this is an arithmetic square root, that is a non-negative number, square root of which is $A$. This is a standard agreement in Algebra. Whenever both positive and negative values are allowed, we usually use the notation $\pm \sqrt{A}$.
This agreement necessitates that the right side of an equation must be non-negative, that is $x \ge 2$

Combination of two restrictions, $x \ge - 10$ and $x \ge 2$ results in one condition on an argument $x$:
$x \ge 2$

With this restriction in mind, let's square the equation:
$x + 10 = {\left(x - 2\right)}^{2}$
After opening the parenthesis it's transformed into
${x}^{2} - 5 x - 6 = 0$
Two roots of this equation are ${x}_{1} = 6$ and ${x}_{2} = - 1$.
Only the first solution satisfy the initial restriction $x \ge 2$. Therefore, the only solution to an original equation is ${x}_{1} = 6$.

CHECK
For ${x}_{1} = 6$ the left side is $\sqrt{6 + 10} = 4$
and the right side is $6 - 2 = 4$
This is an identity $4 = 4$, which confirms that we have solved an equation correctly.