## Boxman61 3 years ago Solve: ___ ______ √x-2 -1 = √2(x-3)

1. MCronin

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

2. Boxman61

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

3. MCronin

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

4. Boxman61

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

5. MCronin

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

6. Denebel

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

7. MCronin

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

8. Boxman61

i appreciate the help

9. MCronin

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

10. MCronin

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

11. Denebel

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

12. Denebel

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.

13. Denebel

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

14. Boxman61

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