# Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 118

$\{6\}$

#### Work Step by Step

For the root to be real, x must be $\geq-10.$ The LHS is nonnegative, so the RHS = $x-2$ is also nonnegative. All solutions must satisfy $x\geq 2\qquad (*)$ $\sqrt{x+10}=x-2 \quad$... Square both sides $x+10=(x-2)^{2}$ $x+10=x^{2}-4x+4$ $0=x^{2}-5x-6$ We factor this trinomial by finding two factors of $-6$ that add to $-5$; they are $-6$ and $+1$. $(x-6)(x+1)=0$ $x=6,\quad x=-1$ $x=-1 \;$ is extraneous (does not satisfy (*) ), so we discard it. $x=6$ is a valid solution Check: $\sqrt{6+10}=6-2$ $\sqrt{16}=4$ The solution set is $\{6\}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.