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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?
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?
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?
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 2x2x3/2x\frac{2x^2-x^{-3/2}}{\sqrt{x}} 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.
Original post by RDKGames
What's the full question??

I'm assuming it is to write 2x2x3/2x\frac{2x^2-x^{-3/2}}{\sqrt{x}} 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.


Untitled.png


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 ?
Reply 6
Original post by samantham999
Untitled.png


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 x12(2x2x32) \displaystyle x^{-\frac{1}{2}}(2x^2-x^{-\frac{3}{2}}) .
Now you just use laws of indices.
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
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 ?
Original post by samantham999
so for 2x^2 over x^1/2 i got 2x^1/2 ?


No.
You have 2122-\frac{1}{2} 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 21/22^{1/2} come from?? How did you get 4/2????

x3212-x^{\frac{3}{2}-\frac{1}{2}} when dividing by root x.
Original post by samantham999
Untitled.png


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
Original post by RDKGames
No.
You have 2122-\frac{1}{2} as the exponent, surely you can do this GCSE problem subtraction.



Where did the 21/22^{1/2} come from?? How did you get 4/2????

x3212-x^{\frac{3}{2}-\frac{1}{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 ?
(edited 7 years ago)
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
(edited 7 years ago)
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.
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 ?
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
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 ?
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
Original post by samantham999
Untitled.png


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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