## anonymous one year ago Solve for x. [(SQRT.x+1)+(SQRT.x-1)]=(SQRT.2x+1)

1. phi

it's not clear what is inside the square root signs. can you use the equation editor?

2. anonymous

$\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1}$

3. anonymous

I know that I need to square everything to get rid of my square roots

4. phi

if you had (a+b) and squared that: (a+b)(a+b) what do you get ?

5. anonymous

$a^{2}+2ab+b ^{2}$

6. phi

now we assume $$a= \sqrt{x+1}$$ and $$b= \sqrt{x-1}$$ in $a^{2}+2ab+b ^{2}$ what do you get ?

7. anonymous

$x ^{2}-1$ ??

8. phi

don't skip steps, just replace a^2, a and b, and b^2 in $a^{2}+2ab+b ^{2}$

9. anonymous

$x ^{2}-x+x-1$

10. phi

a^2 is x+1 b^2 is x-1 2*a*b is $$2 \sqrt{x+1}\sqrt{x-1}$$ so you should get $x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1$

11. phi

we are doing $\left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2$

12. anonymous

Okay....So I can't square $\sqrt{x+1}$ and $\sqrt{x-1}$ individually? I have to foil them together right?

13. phi

I am not following you. yes, you can square $$\sqrt{x+1}$$ to get x+1 but when we square $\left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2$ we follow the same pattern as $(a+b)^2 = a^2 +2ab+b^2$ in other words $\left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2 = x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1$ that simplifies a little: $2x+ 2 \sqrt{x+1}\sqrt{x-1}$

14. phi

meanwhile, we also square the right-hand side $\left( \sqrt{2x+1}\right)^2$

15. anonymous

$4x ^{2}+4x+1$??

16. anonymous

Or would that just actually be 2x+1 ??

17. phi

you did (2x+1)^2 it is easier than that square of a square root: drop the square root sign

18. phi

yes, just 2x+1

19. phi

so now we have $2x+ 2 \sqrt{x+1}\sqrt{x-1} = 2x+1$ I would add -2x to both sides as the next step.

20. anonymous

Okay so $2\sqrt{x+1}\sqrt{x-1} =1$

21. phi

yes. any idea what would be a good step now?

22. anonymous

I know I need to do something with my square roots on the left side, but I don't know what..

23. phi

use this idea: $(a\cdot b)^2 = a^2 \cdot b^2$

24. anonymous

$2\sqrt{x+1}^{2}\sqrt{x-1}^{2}$ ??

25. phi

and also the 2 what you are doing is squaring the whole left side $( 2 \sqrt{x+1} \sqrt{x-1} )^2$ and that means multiply that mess times itself $2 \sqrt{x+1} \sqrt{x-1} \cdot 2 \sqrt{x+1} \sqrt{x-1}$ when you multiply, you can chaange the order $2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1}$

26. phi

the "rule" is if you square (a*b) you square each term inside: a^2 * b^2 (this extends to (a*b*c)^2 or any number of factors: square each one)

27. anonymous

Is this another instance where I can drop my square root sign??

28. phi

yes, $\sqrt{x}\cdot \sqrt{x} =x$ we can use that rule twice in $2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1}$

29. anonymous

So, $4\times(x+1)\times(x-1)$

30. anonymous

Would I distribute my four to the (x+1) and (x-1) and then foil??

31. phi

yes, but we should not use $$\times$$ in algebra (it looks like an x) we also have to square the right-hand side and write the whole equation

32. anonymous

$4[(x+1)*(x-1)] = 1^{2}$ $(4x+4)(4x-4) = 1$ Am I good so far?

33. phi

you only multiply by 4 once the easiest way is to first do (x+1)(x-1) you get x^2 -1 (difference of squares) now multiply by 4 4(x^2-1) = 4x^2 -4 that is all equal to the right side 4x^2-4 = 1

34. phi

FYI, notice when you got (4x+4) (4x-4) that is the same as 4*(x+1) * 4 * (x-1) but we only have 4*(x+1)*(x-1)

35. anonymous

I got $\sqrt{\frac{ 1 }{ 2 }}$

36. phi

starting with $4x^2-4 = 1$ add +4 to both sides

37. anonymous

like $x = \sqrt{\frac{ 5 }{ 2 }}$

38. phi

closing in on it, but still wrong

39. anonymous

oh my goodness. $x = \frac{ \sqrt{5} }{ 4 }$

40. phi

4x^2 = 5 x^2 = 5/4 x= sqrt(5/4) which is also $\frac{\sqrt{5}}{\sqrt{4}}= \frac{\sqrt{5}}{2}$

41. anonymous

Okay got it. Thank you so much!!!

42. phi

we should also check x= - sqrt(5) /2 but we can throw out the negative square root because in the original problem we have $$\sqrt{x-1}$$ and if x is negative, we will have a negative number inside a square root and we don't allow that. in other words, it's a good idea to make a note that x= - sqrt(5)/2 is not allowed.

43. anonymous

I really appreciate your help and patience with me. I am working on a review for a test I have tomorrow and was really struggling with this one. I really appreciate the help!!