Question

tle="R = \sqrt{ \frac{ax - P}{Q + bx} } " alt="R = \sqrt{ \frac{ax - P}{Q + bx} } " align="absmiddle" class="latex-formula">
solve for x. Please can someone help me ASAP. I need to hand it on today.​

Answer 1

Step-by-step explanation:

$r = \sqrt{ \frac{ax - p}{q + bx} } \\ {r}^{2} = \frac{ax - p}{q + bx}$

r² (q + bx) = ax - p

qr² + bxr² = ax - p

qr² + p = ax - bxr²

qr² + p = x (a - br²)

$x = \frac{q {r}^{2} + p}{a - b {r}^{2} }$

Answer 2

Answer:

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

Step-by-step explanation:

$R=\sqrt{\frac{ax-P}{Q+bx}}$

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

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

$R^2=\frac{ax-P}{Q+bx}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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