Solve for x.
[(SQRT.x+1)+(SQRT.x-1)]=(SQRT.2x+1)

- anonymous

Solve for x.
[(SQRT.x+1)+(SQRT.x-1)]=(SQRT.2x+1)

- katieb

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- phi

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

- anonymous

\[\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1}\]

- anonymous

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

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## More answers

- phi

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

- anonymous

\[a^{2}+2ab+b ^{2}\]

- phi

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

- anonymous

\[x ^{2}-1 \] ??

- phi

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

- anonymous

\[x ^{2}-x+x-1 \]

- 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
\[ x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1 \]

- phi

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

- anonymous

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

- 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:
\[ 2x+ 2 \sqrt{x+1}\sqrt{x-1} \]

- phi

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

- anonymous

\[4x ^{2}+4x+1 \]??

- anonymous

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

- phi

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

- phi

yes, just 2x+1

- 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.

- anonymous

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

- phi

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

- anonymous

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

- phi

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

- anonymous

\[2\sqrt{x+1}^{2}\sqrt{x-1}^{2}\] ??

- 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
\[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1} \]

- 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)

- anonymous

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

- phi

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

- anonymous

So, \[4\times(x+1)\times(x-1)\]

- anonymous

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

- 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

- anonymous

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

- 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

- 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)

- anonymous

I got \[\sqrt{\frac{ 1 }{ 2 }}\]

- phi

starting with
\[ 4x^2-4 = 1 \]
add +4 to both sides

- anonymous

like
\[x = \sqrt{\frac{ 5 }{ 2 }}\]

- phi

closing in on it, but still wrong

- anonymous

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

- phi

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

- anonymous

Okay got it. Thank you so much!!!

- 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.

- 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!!

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