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Im currently stuck on this equation I need to modify to be in terms of x

$$y=-x^2+4$$

I got something like this which looks wrong

$$x = -\sqrt{y+4}$$

First you would subtract the 4 from both sides. Do you then divide by -1 then square root it?

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  • $\begingroup$ Solve according to SAMDEB. Solving an equation is the unraveling of an evaluation, so you need to reverse the steps that you would normally use to evaluate an expression. If I gave you instructions on how to get to Wal-Mart, could you find your way back home again? It's the same idea. $\endgroup$ – John Joy Aug 29 '15 at 4:05
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You should divide by $-1$ and then square root it. In the original equation, it is $y = -(x^2) + 4$. Doing this should get you $x = \sqrt{4-y}$.

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    $\begingroup$ Don't forget the $\pm$ $\endgroup$ – abiessu Aug 29 '15 at 3:36
  • $\begingroup$ oh okay thank you very much thats what I was thinking but wasnt sure $\endgroup$ – MD_90 Aug 29 '15 at 3:36
  • $\begingroup$ should it be $$\pm \sqrt{4-y}$$ ? $\endgroup$ – MD_90 Aug 29 '15 at 3:38
  • $\begingroup$ @MD_90, yeah it should have a $\pm$ as abiessu pointed out. I've always thought that the $\pm$ is implied by the radical sign. $\endgroup$ – Mike Pierce Aug 29 '15 at 3:39
  • $\begingroup$ im applying this to a calculus 2 problem so that is gonna be strange working with in a integral for area between two curves that has plus and minus square root $\endgroup$ – MD_90 Aug 29 '15 at 3:44
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Original equation, $y=-(x^2)+4$; Minus $4$ from both sides, $y-4=-x^2;$ Multiply $(-1)$ to both sides, $x^2=-(y-4)$; That gives you, $x^2=4-y$; Square root it, $x=\pm\sqrt {4-y}$ Final ans is $x=\pm\sqrt {4-y}$.

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  • $\begingroup$ Good algebra. Do you have anything to add that hasn't been stated in the other answer? $\endgroup$ – abiessu Aug 29 '15 at 3:48
  • $\begingroup$ I forgot to put the ± sign after square rooting. Thank you very much. $\endgroup$ – Tridip das Aug 29 '15 at 3:56

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