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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!

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  • $\begingroup$ Thank you. I removed the extra square (just a typo). Question still stands! $\endgroup$
    – Chris
    Commented Dec 5, 2021 at 19:45

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Chris
    Commented Dec 5, 2021 at 20:01
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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.

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  • $\begingroup$ 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! $\endgroup$
    – Chris
    Commented Dec 5, 2021 at 20:07

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