Polynomials questions with square root

Question

$$\sqrt{x^2 -7x -2} = x^2 -7x -14$$

Hey, I know the answer is $x=-2, x=9. $ I just don't know how to get to it.

I tried squaring both sides, but I think there's an easier way.

Answer 1

Let $p(x)=x^2-7x-2$. Then your equation becomes $\sqrt{p(x)}=p(x)-12$. So, start by solving the equation $\sqrt y=y-12$. Use the fact that its only solution is $16$.

Answer 2

Hint

First let $a=\sqrt{x^2-7x-2}.$ Note that this means $a\ge0$.

Solve $a=a^2-12$.

Then solve for $x$ from $a^2=x^2-7x-2.$

Answer 3

Another way...

Let $u = x^2 - 7x -2$ so you are looking at $$ \sqrt{u} = u-12 \text{.} $$ This is a quadratic in the variable $\sqrt{u}$. \begin{align*} u - \sqrt{u} - 12 &= 0 \\ (\sqrt{u} - 4)(\sqrt{u}+3) &= 0 \end{align*} For the product of two numbers to be zero, at least one of them is zero, so either $\sqrt{u} = 4$ or $\sqrt{u} = -3$ (or both, but that can't happen here). Since no real number has $-3$ as its square root, we must have $u = 16$. So, $$ x^2 - 7x - 18 = 0 $$ and we can factor this as $(x-9)(x+2) = 0$, giving the solutions you mention.

Answer 4

We need to solve $$x^2-7x-2-\sqrt{x^2-7x-2}-12=0$$ or $$\left(\sqrt{x^2-7x-2}+3\right)\left(\sqrt{x^2-7x-2}-4\right)=0$$ or $$\sqrt{x^2-7x-2}=4,$$ which gives $$x^2-7x-2=16$$ and since for these values of $x$ we have $$x^2-7x-14=4>0,$$ we'll get all roots from the equation: $$x^2-7x-18=0,$$ which gives your answer: $$\{-2,9\}$$