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

1. MCronin Group Title

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 Group Title

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

3. MCronin Group Title

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 Group Title

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

5. MCronin Group Title

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

6. Denebel Group Title

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

7. MCronin Group Title

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

8. Boxman61 Group Title

i appreciate the help

9. MCronin Group Title

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

10. MCronin Group Title

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

11. Denebel Group Title

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 Group Title

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 Group Title

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

14. Boxman61 Group Title

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