Very stuck on Maths past paper question

The expression x²-2mx has a minimum value as x varies.

Find the minimum value of x²-2mx
Give your answer in terms of m.

This is from the june 2003 past paper. I can normally figure these things out with a bit of thought but i'm stumped on this one.

Thanks for your help

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Reply 1

logic123

i know this one i think

rearrange to give m which is m=x(1-x)/2 then work it from there.

Reply 2

tc123

x^2 -2mx = (x-m)^2 -k
x^2 -2mx = (x-m)(x-m)-k
x^2 -2mx = x^2 -2mx+m^2 -k
0 = m^2 -k
k = m^2

There you go!

Reply 3

DAVE 101

Sorry, i'm not saying your wrong, but how did you rearrange without an equals sign?

Reply 4

tc123

That was part a) and if you need the answer for part b) just say. :smile:

Reply 5

logic123

both sides of the equation m8

Reply 6

DAVE 101

Ok thanks, i can see how part 1 works. Surely the answer to part B is 0.

0²-2xmx0=0-0 = 0

Minus numbers would return their negitave quivilants due to the x² and the -.

So 0, in terms of m, is 0m?

And to get that, x=m?

Reply 7

logic123

is hat edexel maths cause it seems a bizARE QUESAtion to have the final answer as 0

Reply 8

DAVE 101

yeh it is edexcel maths.

I don't know, the answer being 0 does seem like a trick, but then I've come to expect tricks from exam boards. I suppose they're looking for confidence in your answer.

My logic above does seem correct though?

Reply 9

kadhumia_flo

I got it...

if we say x=m, we can say that x²-2x² and that is the smallest number possible

so x=m

Reply 10

DAVE 101

kadhumia_flo

I got it...

if we say x=m, we can say that x²-2x² and that is the smallest number possible

so x=m

Erm, how does x=m, exactly? I'm not saying you're wrong, just can't see it myself.

Reply 11

style

ok im realy confused here , can someone find the right answer on the mark scheme and go through it step by step so us dummies can understand? :smile:

Reply 12

Agrippa

In any quadratic ax²+bx+c the minimum point is found where

If you know this, then you proceed as follows:

If you don't know this, then you proceed as follows:
First, complete the square (put the quadratic into the form a(x-p)²+q)

m is fixed, so we can't change the value of m². So we want to make (x-m)² as small as possible. You know that all squares are non-negative, so the smallest value is at (x-m)²=0

Hope this has been of help.

Reply 13

Agrippa

kadhumia_flo

I got it...

if we say x=m, we can say that x²-2x² and that is the smallest number possible

so x=m

It's right, but you wouldn't get the marks on an exam. You need to show that the condition x=m will give the smallest value, not just say so. Or show that having the smallest value implies x=m (what I've done above).

Reply 14

logic123

to be frank judghing by that question, its too hard for our exam baord AQA :smile:

Reply 15

DAVE 101

I was JUST about to post a reply asking how on earth you did that, but i re-read your reply several times, and it see now. Cheers julianparmar, i wouldn't have sussed that by myself

Reply 16

Leonidas

logic123

to be frank judghing by that question, its too hard for our exam baord AQA :smile:

Haha, AQA has its moments too.

Reply 17

style

hold up, was this on a calculator paper :confused:

Reply 18

style

fanx julianparmer ur a star! i get it now but im still praying nothin lyk that cmz up.

Reply 19

kadhumia_flo

How did you know that formula? I'm on Edexcel (modular) and i've never heard of it at all

is that like some a-level formula

if thats the only way, then that question is definately not coming up