## anonymous one year ago need help !! :))

1. anonymous

with?

2. anonymous

3. phi

line 1 is ok. but line 2 is wrong in other words, (x-y)^2 is *not* x^2-y^2 (which is what they did) (x-y)^2 is x^2 +y^2 -2xy

4. anonymous

what am i supposed to d then? how do i solve it

5. phi

You did not post the question. I assume they want to know which line contains a mistake. If so, that would be line 2.

6. anonymous

i have to find the mistake then solve it correctly

7. phi

ok. line 1 is technically not a mistake, but it is the wrong first step (line 2 is definitely a mistake) I would start over, with the original $x - \sqrt{17-4x}= 3$ and put the square root on the right-hand side , like this: $x - \sqrt{17-4x} +\sqrt{17-4x} = 3 + \sqrt{17-4x}$ the left side simplifies $x= 3 + \sqrt{17-4x}$ now add -3 to both sides $x-3 = \sqrt{17-4x}$

8. phi

now square both sides. on the left that means write (x-3)^2 on the right , what do you get after squaring ? any idea?

9. anonymous

X^2-6X+9?

10. phi

yes, that is the left side. the right side is "easy" --- squaring gets rid of the square root side. can you write down the whole equation (both sides)?

11. anonymous

X^2-6X+9=17-4X

12. phi

now add -17 to both sides and +4x to both sides

13. anonymous

X^2-2X-8?

14. phi

you mean X^2-2X-8=0 remember it is an equation

15. anonymous

A=1,B=-2,C=-8

16. phi

yes. you can use the quadratic formula (if they want you to) but you can factor this: notice -4* 2= -8 and -4+2= -2

17. anonymous

OH YEAH

18. anonymous

THATS IT? thanks so much!! :)

19. phi

(x-4)(x+2)= 0 either x-4=0 --> x=4 or x+2 =0 ---> x= -2 however, with these types of problems you have to check both solutions. sometimes you get an "extraneous" solution. So check both values in the original equation

20. anonymous

-2 is false. thanks:)

21. phi

yes