Asked by Alehanda
solve x-sqrt(2x+1)=7
Can someone explain how to work this out. Im not exactly sure how you figure the square root part.
Can someone explain how to work this out. Im not exactly sure how you figure the square root part.
Answers
Answered by
Damon
sqrt(2x+1) = (x-7)
square both sides
2x+1 = x^2 - 14 x + 49
x^2 - 16 x + 48 = 0
(x-4)(x-12) = 0
x = 4 or x = 12 MAYBE because when you square both sides you introduce an ambiguity since -a^2 = a^2 . Therefore check both answers in the original problem.
try x = 4
4 - sqrt (9) = 7
NO 4 -3 = 1, we need 4 + 3
try x = 12
12 - sqrt (25) = 12 -5 = 7 YES, we did it. x = 12 works.
square both sides
2x+1 = x^2 - 14 x + 49
x^2 - 16 x + 48 = 0
(x-4)(x-12) = 0
x = 4 or x = 12 MAYBE because when you square both sides you introduce an ambiguity since -a^2 = a^2 . Therefore check both answers in the original problem.
try x = 4
4 - sqrt (9) = 7
NO 4 -3 = 1, we need 4 + 3
try x = 12
12 - sqrt (25) = 12 -5 = 7 YES, we did it. x = 12 works.
Answered by
Count Iblis
x - 7 = sqrt(2x+1) ------>
(x-7)^2 = |2x+1| ------->
(x - 7)^2 = 2 x + 1 And 2x+1>=0 (1)
Or (x - 7)^2 = -2 x - 1 And 2x+1<0 (2)
(1):
x^2-16x+48 = 0 And 2x+1>=0 ------->
x = 4 or x = 12
(2): No solutions.
The solutions are thus x = 4 or x = 12
(x-7)^2 = |2x+1| ------->
(x - 7)^2 = 2 x + 1 And 2x+1>=0 (1)
Or (x - 7)^2 = -2 x - 1 And 2x+1<0 (2)
(1):
x^2-16x+48 = 0 And 2x+1>=0 ------->
x = 4 or x = 12
(2): No solutions.
The solutions are thus x = 4 or x = 12
Answered by
Count Iblis
I forgot to check if x - 7 is positive. So, as Damon pointed out, x = 4 is not a solution.
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