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leonardo1841 · Nov 14, 2009, 10:34 AM

Make x the subject

((Ax2)/b) + x = d

Unknown008 · Nov 14, 2009, 09:54 PM

$\frac{ax^2}{b} + x = d$

Multiply everything by b to remove fractions;

$\frac{ax^2}{\cancel{b}} \cancel{b} + bx = bd$

Now factorise a

$a(x^2+ \frac{b}{a}x) = bd$

Divide by a:

$\frac{\cancel{a}(x^2+ \frac{b}{a}x)}{\cancel{a}} = \frac{bd}{a}$

Complete the square:

$(x^2+ \frac{b}{a}x) = \frac{bd}{a}$

$(x+ \frac{b}{2a})^2 - \frac{b^2}{4a^2} = \frac{bd}{a}$

Add the b^2/4a^2 to both sides;

$(x+ \frac{b}{2a})^2 \cancel{- \frac{b^2}{4a^2} +\frac{b^2}{4a^2}} = \frac{bd}{a} + \frac{b^2}{4a^2}$

Take square root:

$(x+ \frac{b}{2a})= \pm \sqrt{\frac{bd}{a} + \frac{b^2}{4a^2}}$

Subtract b/2a;

$x+ \cancel{\frac{b}{2a} -\frac{b}{2a}} = \pm \sqrt{\frac{bd}{a} + \frac{b^2}{4a^2}} - \frac{b}{2a}$

$x = \pm \sqrt{\frac{bd}{a} + \frac{b^2}{4a^2}} - \frac{b}{2a}$

Of course, there is a shortcut, but I want to show how you do without a formula.