Mathematics

Mathematics, 23.08.2019 10:20, Jesser18

Square root 6g=(j+y) for g. solve the equation

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Mathematics, 23.07.2019 06:00, minersaysay22

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?

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Mathematics, 28.07.2019 21:00, KingKD

Solve x2 = 12x – 15 by completing the square. which is the solution set of the equation? -square root of 51,-6+ square root of 51) - square root of 21,-6+ square root of 21) - square root of 51, 6+ square root of 51) - square root of 21, 6 + square root of 21) i will mark brainliest for the correct answer

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Mathematics, 02.08.2019 21:00, madisonmoore9716

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?

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Mathematics, 30.10.2019 07:31, makaleyf19

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?

Answers: 1

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