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A-Level Maths [Help] - Surds

So i had this question
Show that
[br]f(x)=(34x)2[br]Showthatf(x)=9x1/2+Ax1/2+B[br]whereAandBareconstants[br][br]f(x) = (3-4\sqrt{x})^2[br]Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B[br]where A and B are constants[br]

I done the first bit - expanding
Which leaves me with
Unparseable latex formula:

\frac{9-24\sqrt{x} +16x}{\sqrt{x}} \[br]



I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root
(edited 7 years ago)
Reply 1
bump - fixed formats
Think about what root x would be if you wrote it as x to a power
Then actually try to divide each term of the numerator by this and see what you get - use power laws
Original post by AdeptDz
So i had this question
Show that
[br]f(x)=(34x)2[br]Showthatf(x)=9x1/2+Ax1/2+B[br]whereAandBareconstants[br][br]f(x) = (3-4\sqrt{x})^2[br]Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B[br]where A and B are constants[br]

I done the first bit - expanding
Which leaves me with
Unparseable latex formula:

\frac{9-24\sqrt{x} +16x}{\sqrt{x}} \[br]



I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root


Divide each term by x\sqrt{x} and remember that x=x1/2\sqrt{x}=x^{1/2} also 1a=a1\frac{1}{a}=a^{-1}

Also f(x) cannot be in that form unless it is f(x)x\displaystyle \frac{f(x)}{\sqrt{x}}
(edited 7 years ago)
924x+16xx \frac{9-24\sqrt{x}+16x} {\sqrt{x}} can be written as: 9x+24xx+16xx \frac{9} {\sqrt{x}} + \frac{-24\sqrt{x}} {\sqrt{x}} + \frac{16x}{\sqrt{x}} . See if that helps you.

If you need some extra help with figuring out A:

Spoiler

Reply 5
Original post by RDKGames
Divide each term by x\sqrt{x} and remember that x=x1/2\sqrt{x}=x^{1/2} also 1a=a1\frac{1}{a}=a^{-1}


Yh thanks, i understand what you said. So okay it saids the answer to:
9x=9x1/2 \frac {9}{\sqrt{x}} = 9x^{-1/2}
So like it is
9/x1/2= 9 / x^{1/2} = or is that basically saying 9/a so would it be 9x1/2 9x^{-1/2}
If that isn't correct then I still don't get it :s-smilie:
(edited 7 years ago)
Reply 6
Original post by BobBobson
924x+16xx \frac{9-24\sqrt{x}+16x} {\sqrt{x}} can be written as: 9x+24xx+16xx \frac{9} {\sqrt{x}} + \frac{-24\sqrt{x}} {\sqrt{x}} + \frac{16x}{\sqrt{x}} . See if that helps you.

If you need some extra help with figuring out A:

Spoiler



Thanks, (@ spoiler) I should know this but I still can't figure out why that is the case

Edit: Ohh i think i understand now, it is basically like saying 2÷2=2 2 \div \sqrt{2} = \sqrt{2} right?
(edited 7 years ago)
Reply 7
And then its like saying 24x÷x 24 \sqrt{x} \div \sqrt{x} is the same as root x divided by root x which is 1? so it would just be 24
(edited 7 years ago)
Original post by AdeptDz
Yh thanks, i understand what you said. So okay it saids the answer to:
9x=9x1/2 \frac {9}{\sqrt{x}} = 9x^{-1/2}
So like it is
9/x1/2= 9 / x^{1/2} = or is that basically saying 9/a so would it be 9x1/2 9x^{-1/2}
If that isn't correct then I still don't get it :s-smilie:


That's correct.
Reply 9
Original post by RDKGames
That's correct.


Okay, thanks.. again

And thank you to everyone above aswell
Original post by AdeptDz
So i had this question
Show that
[br]f(x)=(34x)2[br]Showthatf(x)=9x1/2+Ax1/2+B[br]whereAandBareconstants[br][br]f(x) = (3-4\sqrt{x})^2[br]Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B[br]where A and B are constants[br]

I done the first bit - expanding
Which leaves me with
Unparseable latex formula:

\frac{9-24\sqrt{x} +16x}{\sqrt{x}} \[br]



I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root


You've got the answer but in the wrong format, remember root(x) = x^1/2

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