$s = \frac{p x}{d} (\frac{d}{2} - x)$ $s = (px)/d (d/2 - x)$ Make $x$ $x$ the subject of formula..?

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Answer 1 · 2017-12-04T17:41:22

$x = \frac{- p d \pm \sqrt{{(- p d)}^{2} - 16 p s d}}{4 p}$ $x = (-pd +- sqrt( (-pd)^2 - 16psd))/(4p)$

For starters, notice that your original equation can be simplified to

$s = (px)/color(red)(cancel(color(black)(d))) * color(red)(cancel(color(black)(d)))/2 - (px)/d * x$

$s = (px)/2 - (px^2)/d$

with $d !=0$ .

The fractions present on the right side of the equation have $2d$ as the common denominator, so rewrite the equation as

$s = (px)/2 * d/d - (px^2)/d * 2/2$

$s = (pxd - 2px^2)/(2d)$

Multiply both sides by $2d$ to get

$2sd = pdx - 2px^2$

Rearrange the equation to quadratic form

$2px^2 - pdx + 2sd = 0$

At this point, you can use the quadratic formula to make $x$ the subject of the equation. You know that for a general-form quadratic equation

$color(blue)(ax^2 + bx + c = 0 )$

the quadratic formula looks like this

$color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)$

In your case, you have

$a = 2p$

$b = -pd$

$c = 2sd$

This means that $x$ will be

$x = (-(-pd) +- sqrt( (-pd)^2 - 4 * 2p * 2sd))/(2 * 2p)$

$x = (pd +- sqrt( (-pd)^2 - 16psd))/(4p)$

with $p !=0$ .

s = (px)/d (d/2 - x)s=pxd(d2−x)s = (px)/d (d/2 - x) Make xxx the subject of formula..?