## anonymous one year ago please help !!!!! Choose any four (4) of the five radical equations below. Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous i just need you guys to help me in one question 4. 2√4 - 2x = 2 - x please help !!! how do we solve it ?

1. random231

|dw:1441477010872:dw| now square both sides and solve the eqn

2. anonymous

$2\sqrt{4}-2x=2-x$ Rearrange.$2\sqrt{4}-2=2x-x$$\boxed{x=2\sqrt{4}-2}$

3. IrishBoy123

$\sqrt{4}$ is this for real?!

4. anonymous

No, I think it's four real xD im sorry i had to

5. anonymous

You didnt simplify my answer? :(

6. IrishBoy123

it's "four real" !!! lol!!

7. anonymous

jhannybean, the 2x is also the under the radical. so can we really send it to the other side ?

8. anonymous

so your function is $$2\sqrt{4-2x}= 2-x$$?

9. anonymous

yeah

10. anonymous

Okay. Then we would have to begin by squaring both sides. OUr goal is to isolate the x's to one side.

11. anonymous

$(2\sqrt{4-2x})^2 = (2-x)^2$$4(4-2x) = 4-4x+x^2$$16-8x=4-4x+x^2$$x^2-4x+8x=16-4$$x^2+4x-12=0$Can you factor this now?

12. anonymous

um is it (x + 2)(x - 6) = 0 ?

13. anonymous

$(x+2)(x-6) = x^2 -4x-12 \ne x^2+4x-12$

14. anonymous

oh i think i got it is it (x - 2)( x + 6 ) ?

15. anonymous

Yep

16. anonymous

then we simplify like this : x - 2 = 0 x = 2 and x + 6 = 0 x = - 6 so the answer is x = 2, -6

17. anonymous

Mmhmm

18. anonymous

omg thank you so much :)

19. anonymous

No problem :)

20. anonymous

can i ask u one more question ?

21. anonymous

Sure

22. anonymous

it is similar, $\sqrt{5x - x } = 0$

23. anonymous

how do we solve this ?

24. anonymous

Square both sides first, right?

25. anonymous

Again, we want to isolate x so we need to square both sides. Same method.

26. anonymous

wouldnt it just be like 4x = 0 at the end? xD or just plain out 0?

27. anonymous

You are correct.

28. anonymous

Oooooooo, look at me im doin stuff xD

29. anonymous

oh sorry , made a mistake again i meant $\sqrt{5x} - x = 0$

30. anonymous

Wouldnt that be the same outcome?

31. anonymous

Er... nevermind

32. anonymous

$\sqrt{5x}-x=0$$\sqrt{5x}=x$$(\sqrt{5x})^2=x^2$$5x=x^2$$x^2-5x=0$$x(x-5)=0$$x=0$$x-5=0 \iff x=5$

33. anonymous

@BweadedChicken no, not the same outccome.

34. anonymous

oh that was it ? i thought we need to complete the square and all any way , thank you you're the best :)

35. anonymous

Yeah, completing the square comes along with complex quadratic functions that need to be simplified. You'll know when you see them.

36. anonymous

oh ok :)