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Core 2 Revision - Help!

I am stuck on a disguised quadratic question - part iii) b), below.

I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

Thanks.

Maths Question.png
Reply 1
Original post by alde123
I am stuck on a disguised quadratic question - part iii) b), below.

I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

Thanks.

Maths Question.png

k12k^\frac{1}{2} is the positive square root of kk. So we have

k=5\sqrt{k} = -5

k=25k=25 is not a solution to this. Actually this equation has no solutions.
Reply 2
Original post by Middriver

Spoiler


That doesn't make a difference for this question.
Original post by notnek
That doesn't make a difference for this question.


Why is k^1/2 the positive root and not just the root of k?
Reply 4
Original post by Middriver
Why is k^1/2 the positive root and not just the root of k?

f(x)=x12f(x)=x^\frac{1}{2} is equivalent to the function f(x)=xf(x)=\sqrt{x}.

Both are single valued functions (as all functions are) that return the positive square root of the input.

All positive numbers nn have 2 square roots: n12n^\frac{1}{2} and n12-n^\frac{1}{2} where n12n^\frac{1}{2} is the positive square root of nn.
Reply 5
Thanks notnek and Middriver, but I'm still a bit confused.
Surely when you get k^1/2 = -5 and you square it, you get k = 25 which is > 1 so counts as a +ve solution.

I've put a link to the mark scheme below
http://www.ocr.org.uk/Images/61191-mark-scheme-june.pdf
Its core 2 question 9 part iii) b)
(edited 7 years ago)
Original post by alde123
I am stuck on a disguised quadratic question - part iii) b), below.

I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

Thanks.

Maths Question.png


Wait what exam board are you doing... i thought all Core 2's are over? o.0
Reply 7
Original post by Someboady
Wait what exam board are you doing... i thought all Core 2's are over? o.0


I'm doing OCR but I'm in year 12 - my school's scrapped linear A-levels and is giving us end of year mocks instead.
Original post by alde123
I'm doing OCR but I'm in year 12 - my school's scrapped linear A-levels and is giving us end of year mocks instead.


I don't follow? Maths is still a modular subject as far as I know. You mean they're starting linear A-levels and they're making you do all of A-level Maths over 2 years?
Reply 9
Original post by alde123
Thanks notnek and Middriver, but I'm still a bit confused.
Surely when you get k^1/2 = -5 and you square it, you get k = 25 which is > 1 so counts as a +ve solution

Whenever you a square both sides of an equation, the solutions you obtain are not guaranteed to be correct - you should always check them.

Example : Solve x1=2x-1 = 2

Square both sides :

x22x+1=4x22x3=0(x+1)(x3)=0x=1,x=3x^2-2x+1 = 4 \Rightarrow x^2-2x-3 = 0 \Rightarrow (x+1)(x-3) = 0 \Rightarrow x = -1, x=3


For your equation k12=5\displaystyle k^\frac{1}{2}=-5:

k=25k = 25 is not a solution since 2512=5\displaystyle 25^\frac{1}{2} = 5.
(edited 7 years ago)
Reply 10
Original post by Someboady
I don't follow? Maths is still a modular subject as far as I know. You mean they're starting linear A-levels and they're making you do all of A-level Maths over 2 years?


Yup.
There is still an AS in the current specification right now, but (I don't really know how this works) my school decided that starting with my year group, everyone is going to sit the whole A-Level in Year 13. The only people who sat the their A-Level in maths this year, in my school, were the people doing Further Maths (which I'm not doing).
Not all schools have gone linear though.
(edited 7 years ago)
Original post by alde123
Yup.
There is still an AS in the current specification right now, but (I don't really know how this works) my school decided that starting with my year group, everyone is going to sit the whole A-Level in Year 13. The only people who sat the their A-Level in maths this year, in my school, were the people doing Further Maths (which I'm not doing).
Not all schools have gone linear though.


If you are still stuck after reading Notnek's explanations, have a look at this thread.

What the graph shows is that there is no value k such that k^1/2 = -5.
(edited 7 years ago)
Reply 12
Original post by notnek
Whenever you a square both sides of an equation, the solutions you obtain are not guaranteed to be correct - you should always check them.

Example : Solve x1=2x-1 = 2

Square both sides :

x22x+1=4x22x3=0(x+1)(x3)=0x=1,x=3x^2-2x+1 = 4 \Rightarrow x^2-2x-3 = 0 \Rightarrow (x+1)(x-3) = 0 \Rightarrow x = -1, x=3


For your equation k12=5\displaystyle k^\frac{1}{2}=-5:

k=25k = 25 is not a solution since 2512=5\displaystyle 25^\frac{1}{2} = 5.


Thanks, that makes so much more sense now. :smile:

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