How do you solve for x? √(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 © Algebra House
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