Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 - 3 = x Square root of negative 2x plus 1 - 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 - 1 = x + 3 - 1 Square root of negative 2 x = x + 2 (Square root of negative 2 x) 2 = (x - 4) 2 - 2x = x2 - 8x + 16 - 2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2) (x + 8) x + 2 = 0 x + 8 = 0 x + 2 - 2 = 0 - 2 x + 8 - 8 = 0 - 8 x = - 2 x = - 8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make?

Question

Braiden

Mathematics

5 December, 21:56

Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 - 3 = x Square root of negative 2x plus 1 - 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 - 1 = x + 3 - 1 Square root of negative 2 x = x + 2 (Square root of negative 2 x) 2 = (x - 4) 2 - 2x = x2 - 8x + 16 - 2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2) (x + 8) x + 2 = 0 x + 8 = 0 x + 2 - 2 = 0 - 2 x + 8 - 8 = 0 - 8 x = - 2 x = - 8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make?

+3

Answers (1)

Know the Answer?

Serrano · Answer 1

She kept the + 3 on the right side of the equation in her second step, she forgot to eliminate it ... there fore, it would have been - 2x + 1 = x + 3 not - 2x + 1 - 3 + 3 = x+3