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.
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.
3 answers
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
I forgot to check if x - 7 is positive. So, as Damon pointed out, x = 4 is not a solution.
0 answers