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what's the inverse function of y= sqrt(x^2 + 7x) please help! :)

Mathematics
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\[y=\sqrt(x^2+7x)\]
x(y) = 1/2 (-7±sqrt(4 y^2+49))
so far I have it down to \[x^2=y^2+7y \]

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Other answers:

I'm just kind of confused
\[\large y=\sqrt{x^2+7x} \qquad \rightarrow \qquad x=\sqrt{y^2+7y}\]Squaring both sides,\[\large x^2=y^2+7y\]Let's complete the square on y, half of 7, squared, is \(\dfrac{49}{4}\).\[\large x^2+\frac{49}{4}=y^2+7y+\frac{49}{4}\] \[\large x^2+\frac{49}{4}=\left(y+\frac{7}{2}\right)^2\] I think maybe do something like this to deal with the different degrees of y. Does that make sense?
umm kind of... I'm a little lost at the part where you do the 49/4. This is a homework question online and they want a y= something, equation
Yah there would be a couple more steps to get it there :) ~Taking the square root of both sides,\[\large \pm\sqrt{x^2+\frac{49}{4}}=y+\frac{7}{2}\]
Confused about the `completing the square` step though?
yeah a little
\[y=(\frac{ 1 }{ 2 } (-7±\sqrt{(4 x^2+49)})\]
\[\large x^2+bx\]To complete the square, we take half of the b term, and square it. That is the term we will want to ADD to complete the square.\[\large \left(\frac{b}{2}\right)^2 \qquad \rightarrow \qquad \frac{b^2}{4}\]So to complete the square we would add this term,\[\large x^2+bx+\frac{b^2}{4}\]And since it's a perfect square, we can write it as,\[\large \left(x+\frac{b}{2}\right)^2\]
Hmm completing hte square can be a little tricky if you don't remmeber how to do it :( I can't think of the best way to explain it.
the answer prebz gives is the right one, is that what you got zepdrix?
Yah that's the same :) His just has the 1/2 factored out, so it looks nicer.

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