Exponential problem: perform operation: (x^-1/2+x^1/2)^2

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Violagirl

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Perform the operation:

(x^-1/2+x^1/2)^2

I'm not really sure how to approach this problem. I think that either the 1/2s would cancel out and then 1^2=1 but I'm not sure if this is right.
 
Violagirl said:
Perform the operation:

(x^-1/2+x^1/2)^2
Click to expand...
Do you mean either of the following?

. . . . .(x[sup:1gi0yfm1]-1[/sup:1gi0yfm1])/2 + (x[sup:1gi0yfm1]1[/sup:1gi0yfm1])/2)[sup:1gi0yfm1]2[/sup:1gi0yfm1]

. . . . .(x[sup:1gi0yfm1]-1/2[/sup:1gi0yfm1] + x[sup:1gi0yfm1]1/2[/sup:1gi0yfm1])[sup:1gi0yfm1]2[/sup:1gi0yfm1]

Or something else?

Violagirl said:
I think that either the 1/2s would cancel out...
Click to expand...
Um... With what would the halves "cancel"...? Please clarify. You wrote out the square, multiplied the two binomials, and... then what?

Please be complete. Thank you! :D
 
Sorry! This is the one I meant: .(x-1/2 + x1/2)2

Like would the exponental halves cancel out? So you'd get x^0 which is one. Then one sq. is one. But I have no idea if that's right. =/
 
Violagirl said:
Sorry! This is the one I meant: .(x-1/2 + x1/2)2
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Um... You've now posted "(x - 1/2 + (x1)/2)(2)"...?

It really would help if you would kindly use the formatting explained in the links you saw in the "Read Before Posting" thread that you read before posting. Thank you! :D

Violagirl said:
Like would the exponental halves cancel out? So you'd get x^0 which is one.
Click to expand...
Um... No, there is no rule saying that x[sup:2xkh7xon]m[/sup:2xkh7xon] + x[sup:2xkh7xon]n[/sup:2xkh7xon] somehow equals x[sup:2xkh7xon]m+n[/sup:2xkh7xon]. :shock:

To learn how exponents work, try studying from some of the many great lessons available online! :idea:

. . . . .Google results for "exponent rules"

Have fun! :D

Eliz.
 
Sorry about the confusion! In your first post, I meant the second problem that you stated. I know when you have negative exponents, you're supposed to switch them around. For x^-1/2, I think I'd have to put it into a fraction to make it 1/x^1/2 and then have (1/x^1/2+x^1/2)^2 and then mulitply with a common denominator so they can add up. I'm not sure how to continue on from there.
 
This might help you get started:

Let a = x[sup:2do3vx6i]-1/2[/sup:2do3vx6i]

Let b = x[sup:2do3vx6i]1/2[/sup:2do3vx6i]

Then, (a + b)[sup:2do3vx6i]2[/sup:2do3vx6i] = a[sup:2do3vx6i]2[/sup:2do3vx6i] + 2ab + b[sup:2do3vx6i]2[/sup:2do3vx6i]

Now substiture your x values back in for a and b. Then, simplify.
 
Substiture is a new word meaning "to substitue your". Isn't math fun!!!

Continuing from my previous post,

If a = x[sup:3bsoe9jx]-1/2[/sup:3bsoe9jx] and b = x[sup:3bsoe9jx]1/2[/sup:3bsoe9jx],

then,

(a + b)[sup:3bsoe9jx]2[/sup:3bsoe9jx] would be the same as (x[sup:3bsoe9jx]-1/2[/sup:3bsoe9jx] + x[sup:3bsoe9jx]1/2)[/sup:3bsoe9jx])[sup:3bsoe9jx]2[/sup:3bsoe9jx]

Since we know that (a + b)[sup:3bsoe9jx]2[/sup:3bsoe9jx] = a[sup:3bsoe9jx]2[/sup:3bsoe9jx] + 2ab + b[sup:3bsoe9jx]2[/sup:3bsoe9jx], we can "substiture".

(x[sup:3bsoe9jx]-1/2[/sup:3bsoe9jx])[sup:3bsoe9jx]2[/sup:3bsoe9jx] + 2(x[sup:3bsoe9jx]-1/2[/sup:3bsoe9jx])x[sup:3bsoe9jx]1/2[/sup:3bsoe9jx]) + (x[sup:3bsoe9jx]1/2[/sup:3bsoe9jx])[sup:3bsoe9jx]2[/sup:3bsoe9jx]

Now use your rules of exponents to complete the simplificaton.

x[sup:3bsoe9jx]-1[/sup:3bsoe9jx] + 2x[sup:3bsoe9jx]0[/sup:3bsoe9jx] + x[sup:3bsoe9jx]1[/sup:3bsoe9jx]

Can you finish?
 
I get it now! That makes it a lot easier! So the answer would be x+2+1/x. Thanks!
 
I like how it always takes stapel a while to figure out what someone is trying to say but other people get it in a second.
 
Don't make that mistake. Stapel gets it too ...

She is just trying to teach people to communicate properly - in mathematical terms - which would help them immensely.

I myself do not have that time - so when problem is not stated correctly - even if I can guess it - most of the time I stay away from it. That is a selfish reason - and I commend Stapel for taking up the tough job of teaching the correct grammar for mathematics.
 
Subhotosh Khan said:
Don't make that mistake. Stapel gets it too ...

She is just trying to teach people to communicate properly - in mathematical terms - which would help them immensely.

I myself do not have that time - so when problem is not stated correctly - even if I can guess it - most of the time I stay away from it. That is a selfish reason - and I commend Stapel for taking up the tough job of teaching the correct grammar for mathematics.
Click to expand...

AMEN!
 
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