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Help on these Surds and Indices questions desperately needed!

20160917_130340.jpg
The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!
Reply 1
Original post by dp00
The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!


What have you tried? What are your thoughts?
(edited 7 years ago)
Original post by dp00

The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!


7 - Rationalise the denominator by multiplying top and bottom by 2\sqrt{2}. Proceed to solve for x.

8 - Rationalise the denominator by multiplying top and bottom by the conjugate of 128\sqrt{12}-\sqrt{8} (of course, you can simplify these surds too)
Reply 3
Q7 =5 root 2
I am really sorry, but I just don't understand how you got that answer! Would you mind explaining it and going through the working out?
Reply 5
Original post by dp00
I am really sorry, but I just don't understand how you got that answer! Would you mind explaining it and going through the working out?


No. What have you tried? What are your thoughts, we're here to help you and point you in the right direction, not spoon you the answers. Have you looked at RDK's post? Did you try what he said? What did you get stuck on?
Original post by Zacken
No. What have you tried? What are your thoughts, we're here to help you and point you in the right direction, not spoon you the answers. Have you looked at RDK's post? Did you try what he said? What did you get stuck on?
I am on Q7. I understand that you have to rationalise the denominator, so you multiply 6x by sqrt2 and sqr2 by sqrt2. So you end up with 6xsqrt2 as the numerator and 2 as the denominator. I do not know what to do next...
Reply 7
Original post by dp00
I am on Q7. I understand that you have to rationalise the denominator, so you multiply 6x by sqrt2 and sqr2 by sqrt2. So you end up with 6xsqrt2 as the numerator and 2 as the denominator. I do not know what to do next...


Well, you could do that. But it'd be easier to do this:

ab=c    a=bc\frac{a}{b} = c \iff a = bc.

So here, you'd have 6x2=10+x8\displaystyle \frac{6x}{\sqrt{2}} = 10 + x\sqrt{8} which you can re-arrange to:

6x=2(10+x8)    6x=102+x82=102+x16\displaystyle 6x = \sqrt{2}(10 + x\sqrt{8}) \iff 6x = 10\sqrt{2} + x\sqrt{8}\sqrt{2} = 10\sqrt{2} + x\sqrt{16}
Reply 8
math.pngCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?
Original post by Andrew.G


Sorry for poor quality :smile:


http://www.thestudentroom.co.uk/showthread.php?t=4066671&p=64637319#post64637319

Let's not post full solutions to a question. Thanks! :smile:
Reply 10
Original post by OGFakiie
math.pngCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?


You should make your own thread, not hijack this one, but:

2lnx+1=ln2xlnx2+lne=ln2xx=2e2\ln x + 1 = \ln 2x \Rightarrow \ln x^2 + \ln e = \ln 2x \Rightarrow \cdots \Rightarrow x = \frac{2}{e}
Original post by OGFakiie
math.pngCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?


What have you tried to do??
Original post by Andrew.G
IMG_2557.jpg

Sorry for poor quality :smile:
Thank you very much! That is a really easy method! Could you help me with Q8. For Q8, would you rationalise by multiplying by (sqrt12 +sqrt8 divided by sqrt12+sqrt8)?
Reply 13
Original post by RDKGames


Oh sorry, won't happen again :smile:
okay yeah answer is 2e^-1 , but I got 2-e somehow but all my working looks perfectly fine (to me), where have I gone wrong?
mah.PNG
Reply 15
Original post by OGFakiie
okay yeah answer is 2e^-1 , but I got 2-e somehow but all my working looks perfectly fine (to me), where have I gone wrong?
mah.PNG


lnx+1=ln2\ln x + 1 = \ln 2 deson't mean x+e=2x+e = 2. It means lnex=2\ln ex = 2.

Because lnx+1=ln2    lnx+lne=ln2    lnex=ln2\ln x + 1 = \ln 2 \iff \ln x + \ln e = \ln 2 \iff \ln ex = \ln 2.

It's lna+lnb=lnab\ln a + \ln b = \ln ab not lna+lnb=ln(a+b)\ln a + \ln b = \ln (a+b).
Reply 16
Original post by dp00
Thank you very much! That is a really easy method! Could you help me with Q8. For Q8, would you rationalise by multiplying by (sqrt12 +sqrt8 divided by sqrt12+sqrt8)?


Yes you would!

But first you want to solve the square root of 12 and 8, it will make it easier for!
Original post by Zacken
Well, you could do that. But it'd be easier to do this:

ab=c    a=bc\frac{a}{b} = c \iff a = bc.

So here, you'd have 6x2=10+x8\displaystyle \frac{6x}{\sqrt{2}} = 10 + x\sqrt{8} which you can re-arrange to:

6x=2(10+x8)    6x=102+x82=102+x16\displaystyle 6x = \sqrt{2}(10 + x\sqrt{8}) \iff 6x = 10\sqrt{2} + x\sqrt{8}\sqrt{2} = 10\sqrt{2} + x\sqrt{16}
i think I have worked out q8. Is it sqrt3 + sqrt2??
Reply 18
Original post by dp00
i think I have worked out q8. Is it sqrt3 + sqrt2??


Yes.

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