Given the equation $(a x + b) (a x - b) = 0$ $(ax+b)(ax-b)=0$ , how do you solve for x?

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Answer 1 · 2016-12-21T16:17:10

$x = \pm \frac{b}{a}$ $x=+-b/a$

multiply out and rearrange for $x$ $""x$

$(a x + b) (a x - b)$ $(ax+b)(ax-b)$

this is difference of squares

$a^{2} x^{2} - b^{2} = 0$ $a^2x^2-b^2=0$

$a^{2} x^{2} = b^{2}$ $a^2x^2=b^2$

$x^{2} = \frac{b^{2}}{a^{2}}$ $x^2=b^2/a^2$

$x = \pm \sqrt{\frac{b^{2}}{a^{2}}}$ $x=+-sqrt(b^2/a^2$

$x = \pm \frac{b}{a}$ $x=+-b/a$

Given the equation (ax+b)(ax-b)=0(ax+b)(ax−b)=0(ax+b)(ax-b)=0, how do you solve for x?