How do you solve for x?
√(x + 3) = x - 3
√(x + 3) = x - 3
x + 3 = (x - 3)² {squared both sides}
x + 3 = (x - 3)(x - 3) {when you square a binomial, multiply it by itself}
x + 3 = x² - 6x + 9 {used foil method on right}
x² - 7x + 6 = 0 {subtracted x and subtracted 3 from both sides}
(x - 6)(x - 1) = 0 {factored into two binomials}
x - 6 = 0 or x - 1 = 0 {set each factor equal to 0}
x = 6 or x = 1 {solved each equation for x}
If you put 1 back in, it doesn't check...it is an extraneous soultion
x = 6
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