Question

Solve:
___ ______
√x-2 -1 = √2(x-3)

Answer 1

square both sides to get
(x-2)-2\[\sqrt{x-2}\]+1=2x-6
x-1-2\[\sqrt{x-2}/]=2x-6
2\[\sqrt{x-2}\]=-x+5
\[\sqrt{x-2}\]=-x+5/2
square both sides again to get
x-2=(-x+5/2)^2
x=(-x+5/2)^2+2
x=(x^2-10x+25)/4 +2
4x=x^2-10x+27
x^2-14x+27=0
quadratic from there

Answer 2

\[\sqrt{x-2} -1 = \sqrt{2(x-3)}\]
\[(\sqrt{x-2})^{2} - (1)^{2} = \sqrt{2x-6}\]
\[x-2-1 = 2x-6 \]
\[x-3 = 2x-6\]
\[-x = -3\]
x=3

Answer 3

no you have o square te entire leftside so its (sqrt(x-2)-1)^2 so you get
-2sqrt(x-2)+(x-2)+1=-2sqrt(x-2)+x-1

Answer 4

hmm ok, my instructors answer is 3...trying to figure it out by your method

Answer 5

problem is that you cant square the terms individualy to clear the sqrt

Answer 6

Hmm.. it is x^2-14x-33=0

Answer 7

doing itnot in my head at 2 am i got x^2-8x+12=0 but still not geting to "3"

Answer 8

i appreciate the help

Answer 9

holy hell dropped the 2 on the right side gimme one sec

Answer 10

@denebel has the correct quadratic and it also has one answer that is 3

Answer 11

x+5-2*sqrt (x-2) = 2x
x+5 = 2x + 2*sqrt (x-2)
5 = 2x + 2*sqrt (x-2) - x
5-x = 2*sqrt (x-2)
(5-x)^2 = (2*sqrt (x-2))^2
x^2 - 10x + 25 = 4x - 8
0 = -x^2 + 14x + 33
0 = x^2 - 14x - 33

Answer 12

x=3, x=11. Then you check each value by plugging it back into the original equation. Only x=3 checks out, therefore, x=3 is your answer.

Answer 13

Oops Typo
0 = -x^2 + 14x - 33
0 = x^2 - 14x + 33

Answer 14

your awesome, thank you... now to break it down so i learn it lol