Question

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

Answer 1

this is q sub 1 and two as in charges I assume?\[q_1q_2\]right?

Answer 2

Yes

Answer 3

I've gotten this far and then my algebra broke!

Answer 4

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.

Answer 5

OK, I see. I was stuck at the x^2 coefficients. Thanks!!!

Answer 6

welcome :)