# How do I find the $x$-intercept of $y=-\sqrt {x+2}+3$ by using the more difficult factoring?

Forgive me for not knowing any proper terms, I'm trying to teach myself math on YouTube.

This is what I CAN do: if I want to find the x-intercept of $$y=-\sqrt {x+2}+3$$, I know to replace the y with 0 and to add the square root term to both sides (canceling out the right side). I square both sides (removing the square root) and then solve for x.

$$x = 7$$

I've checked this with an online graphing calculator. But as a challenge and to help me understand further, I wanted to figure this out the difficult way.

This is what I CAN'T figure out, when I square both sides immediately:

$$(0)^2 = (-\sqrt{x+2} + 3)^2$$

$$0 = (-\sqrt{x+2} + 3)(-\sqrt{x+2} + 3)$$

Yeah, from that point on, I have tried to figure this out on paper and simply cannot. Is this futile to try it this way? Thank you!

• Thank you. I removed the extra square (just a typo). Question still stands! Dec 5, 2021 at 19:45

It is futile to try this way, but it actually works. We have $$\begin{equation} 0 = (-\sqrt{x+2}+3)^2 = (x+2)- 6\sqrt{x+2} + 9 \end{equation}$$ But $$\sqrt{x+2} = 3$$, hence $$x + 2 - 18 + 9 = 0$$, hence $$x=7$$, which in turn is indeed a solution of the original equation.

• Yes, I see that extra step I was missing (replacing the square root phrase with the number 3), which is where the futility comes in haha. Thank you, this is exactly what I was looking for to help me just that much more. Dec 5, 2021 at 20:01

I don't believe that squaring the equation $$0=-\sqrt{x+2}+3$$ will get you any closer to the solution $$x=7$$. Instead, you can remove the radical by multiplying by the radical conjugate:

$$0(\sqrt{x+2}+3)=(-\sqrt{x+2}+3)(\sqrt{x+2}+3)$$ $$\Rightarrow 0=-(x+2)+9$$ $$\Rightarrow x=7$$

Note that this process does not introduce any extraneous roots, since $$\sqrt{x+2}+3>0$$ for all $$x$$.

If you still wanted to square your equation first, you could still solve it by multiplying by the new radical conjugate, but this would result in a quadratic, and an extraneous solution.

• Fascinating, that is another way to solving this. Thank you so much, I will try to remember the word "conjugate" (which I've seen before) for future problems. What interested me is how it retains the negative sign, something I struggle with and like to practice using. Thank you again! Dec 5, 2021 at 20:07