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Further mathematics (GCSE) Urgent help is required.

Expand and simplify:
2x^1/2(x^1/2-x^-1/2)
I am not very sure how to solve this.
Thank you in advance fellow TSRians.
Reply 1
Avatar for Notnek
Notnek
21
Original post by Wolfram Alpha
Expand and simplify:
2x^1/2(x^1/2-x^-1/2)
I am not very sure how to solve this.
Thank you in advance fellow TSRians.

Expand as you normally would.

And use the fact that xa×xb≡xa+b\displaystyle x^a \times x^b \equiv x^{a+b}.

Post your working if you get stuck.
add the outer power by the inner power of each while putting a 2 in front , 2x1X^1/2+1/2 - 2x1X^1/2+(-1/2) and simplify it to 2X-2
Original post by notnek
Expand as you normally would.

And use the fact that xa×xb≡xa+b\displaystyle x^a \times x^b \equiv x^{a+b}.

Post your working if you get stuck.


I multiplied 2x^1/2 by x^1/2 which gives 2x. Then 2x^1/2 multiplier by x^-1/2 which gives 2x^0. So would the answer just be 2x + 1?
Reply 4
Avatar for Notnek
Notnek
21
Original post by Wolfram Alpha
I multiplied 2x^1/2 by x^1/2 which gives 2x. Then 2x^1/2 multiplier by x^-1/2 which gives 2x^0. So would the answer just be 2x + 1?

2x2x is correct.

But it should be −2x0-2x^0 instead of 2x02x^0. Can you see why?

And 2x0=2×x0=...2x^0 = 2 \times x^0 = ...
Original post by notnek
2x2x is correct.

But it should be −2x0-2x^0 instead of 2x02x^0. Can you see why?

And 2x0=2×x0=...2x^0 = 2 \times x^0 = ...


Ah, I have just noticed the negative sign. So -2x^0 gives -2 rather than zero. Thus the final answer is 2x-2. I thank you for your time. It is much appreciated.
Reply 6
Avatar for Notnek
Notnek
21
Original post by Wolfram Alpha
Ah, I have just noticed the negative sign. So -2x^0 gives -2 rather than zero. Thus the final answer is 2x-2. I thank you for your time. It is much appreciated.

Correct :smile:
Original post by regan1234567
add the outer power by the inner power of each while putting a 2 in front , 2x1X^1/2+1/2 - 2x1X^1/2+(-1/2) and simplify it to 2X-2


Regan, thank you for your answer.
Original post by notnek
Correct :smile:


I have one more question friend. (3x^1/2 - 2x^-1/2)(2x^1/2 - 3x^-1/2) equals 6x - 13 + 1/6x. Am I correct?
Reply 9
Avatar for Notnek
Notnek
21
Original post by Wolfram Alpha
I have one more question friend. (3x^1/2 - 2x^-1/2)(2x^1/2 - 3x^-1/2) equals 6x - 13 + 1/6x. Am I correct?

Nearly but 1/6x is incorrect.

−2x−12×−3x−12=6x(−12−12)=6x−1=6×x−1=...\displaystyle -2x^{-\frac{1}{2}} \times -3x^{-\frac{1}{2}} = 6x^{\left(-\frac{1}{2} - \frac{1}{2}\right)} = 6x^{-1} = 6 \times x^{-1} = ...
Original post by notnek
Nearly but 1/6x is incorrect.

−2x−12×−3x−12=6x(−12−12)=6x−1=6×x−1=...\displaystyle -2x^{-\frac{1}{2}} \times -3x^{-\frac{1}{2}} = 6x^{\left(-\frac{1}{2} - \frac{1}{2}\right)} = 6x^{-1} = 6 \times x^{-1} = ...


I thought 6x^-1 was 1/6x. I am clearly wrong. Could you please enlighten me?
Reply 11
Avatar for Notnek
Notnek
21
Original post by Wolfram Alpha
I thought 6x^-1 was 1/6x. I am clearly wrong. Could you please enlighten me?

Both are wrong but It's not clear if 1/6x means 16x\frac{1}{6}x or 16x\frac{1}{6x}.


Explanation :

You know that x−1=1x\displaystyle x^{-1} = \frac{1}{x}

So 6x−1=6×x−1=6×1x=61×1x=6x\displaystyle 6x^{-1} = 6 \times x^{-1} = 6 \times \frac{1}{x} = \frac{6}{1} \times \frac{1}{x} = \frac{6}{x}


It will be useful for you to become familiar with this rule : ax−n=axn\displaystyle ax^{-n} = \frac{a}{x^n}

E.g. 6x−1=6x\displaystyle 6x^{-1} = \frac{6}{x} or 4x−3=4x3\displaystyle 4x^{-3} = \frac{4}{x^3}.
Original post by notnek
Both are wrong but It's not clear if 1/6x means 16x\frac{1}{6}x or 16x\frac{1}{6x}.


Explanation :

You know that x−1=1x\displaystyle x^{-1} = \frac{1}{x}

So 6x−1=6×x−1=6×1x=61×1x=6x\displaystyle 6x^{-1} = 6 \times x^{-1} = 6 \times \frac{1}{x} = \frac{6}{1} \times \frac{1}{x} = \frac{6}{x}


It will be useful for you to become familiar with this rule : ax−n=axn\displaystyle ax^{-n} = \frac{a}{x^n}

E.g. 6x−1=6x\displaystyle 6x^{-1} = \frac{6}{x} or 4x−3=4x3\displaystyle 4x^{-3} = \frac{4}{x^3}.


Ah, I see! I was aware of that rule but I suppose I got confused when applying it. Mhmm, thanks for your help again.

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