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1. given that 2x^2 - x^2/3 can be written in the form 2x^p - x^q, find values for p & q

Okay, so I split the equation in half and the denominator is the √x

2x^2 ÷ √x and -x^3/2 ÷ √x

Now I use the laws of indices. So √x is x^1/2

I have a feeling this is where I went wrong, is it -1/2?
2. (Original post by samantham999)
given that 2x^2 - x^2/3 can be written in the form 2x^p - x^q, find values for p & q...
Right off the bat the question doesn't lead anywhere unless I'm misinterpreting it. It's already in that form; p=2 and q=2/3. Where did the division by root x come from?
3. (Original post by RDKGames)
Right off the bat the question doesn't lead anywhere unless I'm misinterpreting it. It's already in that form; p=2 and q=2/3. Where did the division by root x come from?
crap, I split the fraction into two different ones with a common denominator? the question asks write down the value of p & q

so i did 2x^2 over root x?
-x^3/2 over root x?
4. (Original post by samantham999)
crap, I split the fraction into two different ones with a common denominator? the question asks write down the value of p & q

so i did 2x^2 over root x?
-x^3/2 over root x?
What's the full question??

I'm assuming it is to write then yeah root of x is x to the power of a half. It cannot be negative since then it would be 1 over root x.
5. (Original post by RDKGames)
What's the full question??

I'm assuming it is to write then yeah root of x is x to the power of a half. It cannot be negative since then it would be 1 over root x.

So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
6. (Original post by samantham999)

So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
The expression in the question can be written as the following .
Now you just use laws of indices.
7. (Original post by samantham999)

So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
Yes
8. (Original post by RDKGames)
Yes

so for 2x^2 over x^1/2 i got 2x^1/2 ?

-x^3/2 over 2^1/2 i got 4/2 ? which is 2/1 ?
9. (Original post by samantham999)
so for 2x^2 over x^1/2 i got 2x^1/2 ?
No.
You have as the exponent, surely you can do this GCSE problem subtraction.

-x^3/2 over 2^1/2 i got 4/2 ? which is 2/1 ?
Where did the come from?? How did you get 4/2????

when dividing by root x.
10. (Original post by samantham999)

So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
yes
11. (Original post by RDKGames)
No.
You have as the exponent, surely you can do this GCSE problem subtraction.

Where did the come from?? How did you get 4/2????

when dividing by root x.

but why isn't it 2x^2 divide x^1/2 ? don't i do 2 - 1/2 like m-n ? i got 2x^3/2 ?
12. (Original post by samantham999)
but why isn't it 2x^2 divide x^1/2 ? don't i do 2 - 1/2 like m-n ? i got 2x^3/2 ?
Yeah it's that for the exponent. And 2-1/2 is not equal to 1/2 as you said first
13. Sqrt(x) is the same as x^(1/2)
Dividing by x^(1/2) is the same as multiplying by x^(-1/2)

So you end up with (2x^2 - x^(2/3)) * x^(-1/2)

Then literally all you have to do is expand the brackets, remembering this law of indices:

x^a * x^b = x^(a+b).

Just in case you haven't got it yet.
14. (Original post by RDKGames)
Yeah it's that for the exponent. And 2-1/2 is not equal to 1/2 as you said first
for -x^3/2 - 1/2 why do i keep getting 4/2 ? which is 2/2 which is 1 ?
15. (Original post by samantham999)
for -x^3/2 - 1/2 why do i keep getting 4/2 ? which is 2/2 which is 1 ?
lol

"4/2 which is 2/2" no it isn't.

You should be getting 2/2 which is 1 because 3/2-1/2=2/2, dunno where you're pulling the 4 from
16. (Original post by RDKGames)
lol

"4/2 which is 2/2" no it isn't.

You should be getting 2/2 which is 1 because 3/2-1/2=2/2, dunno where you're pulling the 4 from
wow i really need to take care, so many stupid mistakes. i got 2/2 which is 1 so its -x^1 ?
17. (Original post by samantham999)
wow i really need to take care, so many stupid mistakes. i got 2/2 which is 1 so its -x^1 ?
Yes
18. (Original post by samantham999)

So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
(2x^2)/(x^1/2)=2x^3/2 because you just take the powers away from each other when yiou divide i.e. 2-1/2=1.5 which is 3/2.
Then you do the same for the other side:
(-x^3/2)/(x^1/2)=-x^2/2 2/2=1
Therefore the ful answer is :
2x^3/2-x^1

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