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

1 Answer
Aug 1, 2015

Answer:

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

Explanation:

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#