Would some one walk me through the steps to get this into quadratic form? I think I am over thinking it. q[2]/(-L+x)^2 = q[1]/x^2

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Would some one walk me through the steps to get this into quadratic form? I think I am over thinking it. q[2]/(-L+x)^2 = q[1]/x^2

Mathematics
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this is q sub 1 and two as in charges I assume?\[q_1q_2\]right?
Yes
I've gotten this far and then my algebra broke!

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Cross-multiply\[\frac{q_2}{(-L+x)^2}=\frac{q_1}{x^2}\to q_2x^2=q_1(-L+x)^2\]expand the second term\[q_2x^2=q_1(L^2-2Lx+x^2)= q_1L^2-2q_1Lx+q_1x^2\]gather like terms all on one side of the equals sign\[q_1x^2-q_2x^2-2q_1Lx+q_1L^2=0\]factor the x^2 to make the coefficient explicit:\[(q_1-q_2)x^2-2q_1Lx+q_1L^2=0\]now you should be able to identify a, b, and c for the quadratic formula.
OK, I see. I was stuck at the x^2 coefficients. Thanks!!!
welcome :)

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