(3x + 7)^1/2 = x  1 3x + 7 = (x  1)² {squared both sides to eliminate the square root sign} 3x + 7 = (x  1)(x  1) {to square a binomial, multiply it by itself} 3x + 7 = x²  2x + 1 {used foil method} x²  5x  6 = 0 {subtracted 3x and 7 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} x = 6 1 is an "extraneous solution". If you substitute 1 back in for x, it does not make a true statement. Therefore, it is an "extraneous solution". © Algebra House
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