How do you solve $x^2-4=0$ ?

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Answer 1 · 2015-08-01T23:30:01

$x_(1,2) = +-2$

The easiest way to solve this quadratic equation is to isolate $x^2$ on one side of the equation, then take the square root of both sides.

To do that, add $4$ to both sides of the equation

$x^2 - color(red)(cancel(color(black)(4))) + color(red)(cancel(color(black)(4))) = 4$

$x^2 = 4$

Taking the square roots of both sides of the equation will get you

$sqrt((x^2)) = sqrt(4)$

$= +-2 = {(x_1 = color(green)(-2)), (x_2 = color(green)(2)) :}$

Alternatively, you could recognize that you're dealing with the difference of two squares, for which you know that

$color(blue)(a^2 - b^2 = (a-b)(a+b)$

In your case, you would get

$x^2 - (2)^2 = 0$

$(x-2)(x+2) = 0$

This equation equals zero if $(x-2)=0$ or if $(x+2) = 0$ , which means that you have

$x+2 = 0 => x_1 = -2$

and

$x-2 = 0 => x_2 = 2$

How do you solve x^2-4=0x^2-4=0?