<p>The question asks to find the inverse of:</p>
<p>sqrt(x^2+2x) with x>0.</p>
<p>I don't know... I've been stumped on this one, as I seem to go around in circles. Any help would be appreciated.</p>
<p>hah, i am so bored b/c i haven't left for college yet that i'll try to answer this for u (i took calc my senior year, which was last year).</p>
<p>take that first equation, then attempt to solve for x</p>
<p>you get x^2 +2x = y^2</p>
<p>then, this is the creative part.. i know there is a name for it but i forget~</p>
<p>go like this</p>
<p>(x + 1)^2 = y^2 + 2</p>
<p>you see how that makes sense? its called completing the square, i think.</p>
<p>then take the square root of both sides, then subtract 1 from both sides to get:</p>
<p>x = sqrt(y^2 +2) -1</p>
<p>then, switch the x and the y, because that's exactly what an inverse does.. just switchin' up the x and y coordinates.</p>
<p>so, y = sqrt(x^2 +2) -1</p>
<p>that works on the domain x>0</p>
<p>there ya go :) [wow i just impressed myself.. its a wonder what a clear mind can do for you.. i think i'll miss this summer (though if i got it wrong, i'll really look like an ass)]</p>
<p>If y = sqrt(x^2 + 2x)
y^2 = x^2 + 2x</p>
<p>Quick solution:
y^2 + 1 = x^2 + 2x + 1 = (x+1)^2
sqrt(y^2 + 1) = x+1
or x = sqrt(y^2+1) - 1</p>
<p>More 'traditional' solution:
x^2 + 2x - y^2 = 0 ;now solve as a quadratic equation
x = {-2 +/- sqrt( 2^2 - (4)(1)(-y^2))} / 2
= {-2 +/- 2 sqrt(1 + y^2)}/2
= -1 +/- sqrt(1+y^2)
Since x > 0, we must have
x = -1 + sqrt(1+y^2)</p>
<p>Thank you so much. I was mulling whether or not I should have completed the square.</p>
<p>i think i screwed mine up a little -- its tough to do this on the internet. then again, i'm terrible at explaining IRL, and i can't even read what i write on my math tests to argue with teachers half the time, even tho i do very well :)</p>