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Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make? O She added 2x after squaring both sides. O She subtracted 1 before squaring both sides. O She factored x+ 10x + 16 incorrectly. She did not check for extraneous solutions

Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make? O She added 2x after squaring both sides. O She subtracted 1 before squaring both sides. O She factored x+ 10x + 16 incorrectly. She did not check for extraneous solutions

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make? O She added 2x after squaring both sides. O She subtracted 1 before squaring both sides. O She factored x 10x + 16 incorrectly. O She did not check for extraneous solutions Math re to search
Transcribed Image Text:Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make? O She added 2x after squaring both sides. O She subtracted 1 before squaring both sides. O She factored x 10x + 16 incorrectly. O She did not check for extraneous solutions Math re to search
Eloise started to solve a radical equation in this way: -2x+1-3=x V-2x+1-3+3 = x +3 -2.x+1 =x+ 3 -2x+1-1= x + 3-1 V-2x = x+ 2 (-2x) = (x - 4)? -2x x2- 8x + 16 -2x +2x = x + 8x + 16 + 2x 0 = x2 + 10x + 16 0 (x + 2)(x+ 8) X +2 = 0 X +2-2 0-2 X + 8 = 0 X + 8 - 8 = 0-8 x = -8 x = -2 Both solutions are extraneous because they don't satisfy the original equation. Ma
Transcribed Image Text:Eloise started to solve a radical equation in this way: -2x+1-3=x V-2x+1-3+3 = x +3 -2.x+1 =x+ 3 -2x+1-1= x + 3-1 V-2x = x+ 2 (-2x) = (x - 4)? -2x x2- 8x + 16 -2x +2x = x + 8x + 16 + 2x 0 = x2 + 10x + 16 0 (x + 2)(x+ 8) X +2 = 0 X +2-2 0-2 X + 8 = 0 X + 8 - 8 = 0-8 x = -8 x = -2 Both solutions are extraneous because they don't satisfy the original equation. Ma
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