## janessia Group Title need help! 2 years ago 2 years ago

1. janessia

$x - 5\sqrt{x} +6=0$

2. Wired

Like the last one, get 5sqrt(x) by itself on one side of the equation, and then square both sides.

3. timo86m

wired is right :)

4. janessia

lol well so far i got $x-6=5\sqrt{x}$

5. janessia

and idk where to go from there

6. timo86m

divide both sides by 5 :)

7. janessia

but i thought you had to square something or something like that

8. timo86m

isolated for sqrt(x) you not there yet div by 5 first

9. Wired

You can do it either way actually. I find it simpler to square both sides at that point so you don't have to deal with fractions. Those fractions would cancel out later anyway, so no need to introduce them in the first place.

10. timo86m

that is also true :)

11. Wired

$(x-6)^{2} = (5\sqrt{x})^{2}$

12. janessia

how do i do $(5\sqrt{x})^2$

13. timo86m

well (a*b) ^n is a^n*b^n right?

14. janessia

yes so the first is 25 and i still dont kno the second part

15. timo86m

25*x :) sqrt(x)^2 is x :)

16. janessia

oh lol im dumb lol

17. timo86m

no you not :)

18. janessia

ha ha well i think so but thanks anyway :)

19. timo86m

are not :P now finish the q :)

20. janessia

so i got$x-36=25x$

21. janessia

?????

22. timo86m

that dont look right

23. janessia

oh wait i did it wrong sorry lol

24. janessia

x-36=25x???

25. timo86m

(x−6)^2 is (x-6)*(x-6) use foil or whatever method you prefer.

26. janessia

27. Wired

Just as an FYI, here's another way to look at squaring a square root: Note that: $\LARGE (X^{S})^{T} = X^{ST}$ $\LARGE \sqrt{x} = x^{1/2}$ $\LARGE (\sqrt{x})^{2} = (x^{1/2})^{2}=x^{(1/2)(2)} = x^{1}$

28. janessia

does this thing have no solution!!??? lol cuz thts wat it seems like!

29. janessia

ohk thnk u

30. Wired

$x-6=5\sqrt{x}$ $(x-6)^{2}=(5\sqrt{x})^{2}$ $(x-6)(x-6)=25x$ $x(x-6)-6(x-6)=25x$

31. Wired

Do you see what to do now?

32. janessia

distribute?

33. Wired

Yep