Question

`R = \sqrt{ (ax - P)/(Q + bx) }`

solve for x. Please can someone help me ASAP. I need to hand it on today.​

Answer 1

`\displaystyle x=\frac{-P-\math{R}^2Q}{\math{R}^2b-a}`

Explanation:

`R=\sqrt{(ax-P)/(Q+bx)}`

`\mathrm{Square\:both\:sides}`

`R^2=\left(\sqrt{(ax-P)/(Q+bx)}\right)^2`

`R^2=(ax-P)/(Q+bx)`

`\mathrm{Multiply\:both\:sides\:by\:}Q+bx`

`\math{R}^2\left(Q+bx\right)=(ax-P)/(Q+bx)\left(Q+bx\right)`

`\math{R}^2\left(Q+bx\right)=ax-P`

`\math{R}^2Q+\math{R}^2bx=ax-P`

`\mathrm{Subtract\:}\math{R}^2Q\mathrm{\:from\:both\:sides}`

`\math{R}^2Q+\math{R}^2bx-\math{R}^2Q=ax-P-\math{R}^2Q`

`\math{R}^2bx=ax-P-\math{R}^2Q`

`\mathrm{Subtract\:}ax\mathrm{\:from\:both\:sides}`

`\math{R}^2bx-ax=ax-P-\math{R}^2Q-ax`

`\math{R}^2bx-ax=-P-\math{R}^2Q`

`\mathrm{Factor}\:\math{R}^2bx-ax`

`x\left(\math{R}^2b-a\right)=-P-\math{R}^2Q`

`\mathrm{Divide\:both\:sides\:by\:}\math{R}^2b-a`

`\frac{x\left(\math{R}^2b-a\right)}{\math{R}^2b-a}=-\frac{P}{\math{R}^2b-a}-\frac{\math{R}^2Q}{\math{R}^2b-a}`

`x=\frac{-P-\math{R}^2Q}{\math{R}^2b-a}`

Answer 2

Explanation:

`r = \sqrt{ (ax - p)/(q + bx) } \\ {r}^(2) = (ax - p)/(q + bx)`

r² (q + bx) = ax - p

qr² + bxr² = ax - p

qr² + p = ax - bxr²

qr² + p = x (a - br²)

`x = \frac{q {r}^(2) + p}{a - b {r}^(2) }`