## anonymous 5 years ago $x = \sqrt{8x}$

1. amistre64

this one right?

2. anonymous

yup

3. amistre64

the answer should be 0 :) but lets see if we can get to that

4. anonymous

kk

5. amistre64

we know to square it to get rid of the squareroot right?

6. anonymous

yeah

7. amistre64

$x^2 = \sqrt{8x}^2$ $x^2 = 8x$

8. anonymous

so 16 + 2sgr = o

9. amistre64

we move it all to one side now right?

10. anonymous

yes.. was mine right?

11. amistre64

$x^2 -8x = 8x -8x$ $x^2-8x = 0$

12. anonymous

ohh

13. amistre64

16 + 2sqr....no

14. anonymous

i hate math D:

15. amistre64

:) itll get easier

16. amistre64

We can factor out an 'x' from our equation now right?

17. anonymous

yeah now that imma have to repeat algebra 1 :(

18. anonymous

yes

19. amistre64

$x(x-8) =0$

20. anonymous

xsqr - 8x = 0

21. amistre64

now the key to solve this is to know that anything times 0 equals 0 right? so we have 2 options... if x = 0 then; (0)(0-8) = 0 and if x = 8 then; (8)(8-8) = (8)(0) = 8 So we have to test for x = 0 and x = 8 in the original problem to see if these are good....

22. anonymous

so x=0 or x=8

23. amistre64

$0=\sqrt{8*0}$ thats true $8 = \sqrt{8*8} = \sqrt{64}$and thats tru; so yes x = 0 or 8

24. anonymous

ok

25. anonymous

x=$\sqrt{42-x}$

26. amistre64

gonna have to post that for someone else to help with; i gotta get to class :)

27. anonymous

thanks

28. anonymous

$\sqrt{8 x}=2 \sqrt{2} \sqrt{x}$$x=2 \sqrt{2} \sqrt{x}$x = 0 or 8