Solve a fractional equation

Question

How do you go from this step:

$$ \frac{a_2b_2}{\left(b_2+x\right)^2}-\frac{a_1b_1}{\left(b_1-x\right)^2} = 0 $$ To this step:

$$ (a_1b_1-a_2b_2)x^2 + 2b_1b_2(a_1+a_2)x + b_1b_2(a_1b_2 - a_2b_1) = 0 $$

Have tried to do it but my maths are a bit rusty at the moment, It would be nice if somebody could do a step by step on how to get to that.

What I have tried: $$ \frac{(a_2b_2(b_1-x)^2)-(a_1b_1(b_2+x)^2)}{\left(b_2+x)^2(b_1-x\right)^2} = 0 $$ multiply denominators and then cross-multiply numerator with denominator, unfortunately I do not know how to continue this.

Answer 1

Assuming that $(b_2+x)^2(b_1-x)^2\ne 0$ we have

$$ (b_2+x)^2(b_1-x)^2\left(\frac{a_2b_2}{\left(b_2+x\right)^2}-\frac{a_1b_1}{\left(b_1-x\right)^2}\right)=(b_1-x)^2a_2b_2-(b_2+x)^2a_1b_1=\\ (b_1^2-2b_1 x+x^2)a_2b_2-(b_2^2+2b_2 x+x^2)a_1b_1=0 $$

the last step is left.

Answer 2

$$ \frac{a_2b_2}{\left(b_2+x\right)^2}=\frac{a_1b_1}{\left(b_1-x\right)^2} $$

Without loss of generality, assume both $a_1b_1$ and $a_2b_2$ are positive, then we take a square root on both sides

$$ \frac{\sqrt{a_2b_2}}{|b_2+x|}=\frac{\sqrt{a_1b_1}}{|b_1-x|} $$

we get

$$ \sqrt{a_2b_2}\cdot(b_1-x)=\pm\sqrt{a_1b_1}\cdot(b_2+x) $$

Can you proceed from here?

Answer 3

What I have tried: $$ \frac{(a_2b_2(b_1-x)^2)-(a_1b_1(b_2+x)^2)}{\left(b_2+x)^2(b_1-x\right)^2} = 0 $$

You are almost there. You have an equation of the form: $\frac{y}{z} = 0$. This implies $y=0$ i.e.

$$ a_2b_2(b_1-x)^2-a_1b_1(b_2+x)^2 =0.$$

Now expand and re-arrange this into the form

$$Ax^2 + Bx + C = 0,$$

and you're done.