Manipulate the Physics Equation $P = I^2R$ to get R by itself

Question

Given that $P = (V^2 R_1)/(R_1 + R_2)^2$, manipulate the equation so that we get $R_1$ by itself and that we have a quadratic equation.

Where $V, P, R_1$, and $R_2$, are variables.

I'm stuck when I find that $R_1 = P(R_1 + R_2)^2/(V^2)$

Answer 1

$P (R_1+R_2)^2 = V^2 R_1$, then you have $P R_1^2 +(2 P R_2- V^2) R_1 + P R_2^2 = 0$, or $R_1^2 + (2R_2-{V^2 \over P}) R_1+ R_2^2 = 0$, which has solutions $R_1 = {1 \over 2} ({V^2 \over P}- 2 R_2) \pm {V^2 \over P}\sqrt{1-4 {PR_2^2 \over V^2}})$.

Answer 2

In general, if $a = bx/(x + c)^2$, then $a(x^2 + 2cx + c^2) = bx$, then $ax^2 + (2ac - b)x + ac^2 = 0$. Now you can use the quadratic formula.

Answer 3

$$P=\frac{{V^2}\cdot R_1}{(R_1+R_2)^2}$$ $$P(R_1+R_2)^2=V^2R_1$$ $$PR_1^2+2PR_1R_2+PR_2^2=V^2R_1$$ $$PR_1^2+2PR_1R_2-V^2R_1+PR_2^2=0$$ $$PR_1^2+R1(2PR_2-V^2)+PR_2^2=0$$

Plug into the quadratic formula, where $a=c=P$, and $b=2PR_2-V^2$.