The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Indices!

Expand 3x^2/3(2x^2/3+2x^-2/3)

HELP MEEEE
Reply 1
Avatar for KaylaB
KaylaB
20
Original post by stephielola
Expand 3x^2/3(2x^2/3+2x^-2/3)

HELP MEEEE


First of all, the way you've typed it is ambiguous. Does you mean
3x23(2x23+2x−23) \dfrac{3x^{2}} {3} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
Or
3x23(2x23+2x−23) \dfrac{3x^{2}} {3(2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})}

I suspect the first one, but anyway, what have you tried already? :h:
Reply 2
Avatar for stephielola
stephielola
OP
7
Original post by KaylaB
First of all, the way you've typed it is ambiguous. Does you mean
3x23(2x23+2x−23) \dfrac{3x^{2}} {3} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
Or
3x23(2x23+2x−23) \dfrac{3x^{2}} {3(2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})}

I suspect the first one, but anyway, what have you tried already? :h:


The first except the root is 2/3 x
Reply 3
Avatar for KaylaB
KaylaB
20
Original post by stephielola
The first except the root is 2/3 x


Ah so this?
3x23(2x23+2x−23) 3x^{\frac{2} {3}} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})

Spoiler



So how have you attempted this already? :h:
Reply 4
Avatar for stephielola
stephielola
OP
7
Original post by KaylaB
Ah so this?
3x23(2x23+2x−23) 3x^{\frac{2} {3}} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})

Spoiler



So how have you attempted this already? :h:


Ahaha!! Yeah sorry about that :tongue: I have tried multiplying out the bracket using the in-dice law. But I am not sure if I am doing it right :s-smilie:
Reply 5
Avatar for KaylaB
KaylaB
20
Original post by stephielola
Ahaha!! Yeah sorry about that :tongue: I have tried multiplying out the bracket using the in-dice law. But I am not sure if I am doing it right :s-smilie:


Care to post your working out so I can see if you're heading in the right direction? :h:
Reply 6
Avatar for stephielola
stephielola
OP
7
Original post by KaylaB
Care to post your working out so I can see if you're heading in the right direction? :h:


I got something like, 6x^2 and an indice of 4/3 +6x^2 and indice to the 0
Reply 7
Avatar for KaylaB
KaylaB
20
Original post by stephielola
I got something like, 6x^2 and an indice of 4/3 +6x^2 and indice to the 0


You are correct about the first term :yep:

With the second term, you are correct in saying that the index would be 0 so it would be 6x0 6x^{0} which is the same as what?

I don't see where you got 6x2 6x^{2} from? :hmmmm: The answer should only have 2 terms :h:
Reply 8
Avatar for stephielola
stephielola
OP
7
Original post by KaylaB
You are correct about the first term :yep:

With the second term, you are correct in saying that the index would be 0 so it would be 6x0 6x^{0} which is the same as what?

I don't see where you got 6x2 6x^{2} from? :hmmmm: The answer should only have 2 terms :h:


Ahh okkkk xx

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you