## anonymous 3 years ago can anyone help me to the answer for this... (will post equation in a sec...)

1. anonymous

$\sqrt{\sqrt{x-2}+x}=2$

2. anonymous

Square both sides

3. anonymous

root(x-2) + x =4 root(x-2) = 4-x square again

4. anonymous

x-2 = 16 - 8x + x^2 9x = 18 + x^2 x^2 - 9x + 18 = 0

5. anonymous

(x-6)(x-3)=0 x=6 or x=3

6. anonymous

@niravshah08 how on earth did you get the first line x-2 = 16 - 8x + x^2 ???

7. anonymous

|dw:1341261837727:dw|

8. anonymous

okay i understand.... but is it wrong if i do this.....

9. anonymous

|dw:1341262013434:dw|

10. anonymous

and if it is wrong please explain why

11. anonymous

but there cant be two different ways of doing it... i think yours is right... hmm

12. slaaibak

Michelle, your method is wrong, because: $(a + b)^2 \neq a^2 + b^2$

13. slaaibak

So, when squaring an equation, you literally put the whole LHS and RHS in huge brackets and square the whole bracket

14. anonymous

so what your saying is that i can do this:

15. slaaibak

Also, at the end, always check your answers, because (usually when there's a square root) sometimes they yield a negative number under the square root, so the answer is not real

16. anonymous

|dw:1341262893119:dw|

17. anonymous

yeah when i did it my method i got a random answer but did notice the - :) just my method is wrong LOL

18. slaaibak

Yeah that's correct. Although, it would be easier to, before you square it, take the x over to the other side. It just makes it easier to solve.

19. anonymous

ah yeah i see i just tried it and it is possible but just involves alot of roots... and i dont like roots... they scare me... D': thanks for the explanation :D lifesaver :)

20. slaaibak

haha yeah, it gets a bit messy.. it's a pleasure :)