How do you solve ${(x + a)}^{2} - b^{2} = 0$ $(x+a)^2-b^2 = 0$ by factoring?

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Answer 1 · 2015-08-28T14:06:44

$x = - a - b or x = - a + b$ $color(green)(x = -a - b or x = -a + b$

We use the identity $p^{2} - q^{2} = (p + q) \cdot (p - q)$ $color(blue)(p^2 - q^2 = (p+q)*(p-q)$

Here $p = x + a$ $p = x+ a$ and $q = b$ $q = b$

Hence ${(x + a)}^{2} - b^{2} = 0$ $(x+a)^2-b^2 = 0$ can be written as :

$(x + a + b) (x + a - b) = 0$ $(x+a + b) (x + a - b) = 0$

This implies that:

$x + a + b = 0 or x + a - b = 0$ $x + a + b = 0 or x + a - b = 0$

$x = - a - b or x = - a + b$ $color(green)(x = -a - b or x = -a + b$

How do you solve (x+a)^2-b^2 = 0(x+a)2−b2=0(x+a)^2-b^2 = 0 by factoring?