The Student Room Group

Completing the square

8x-x^2 can be written in the form p-(x-q)^2, for all values of x:

a) Find value of p and q

b) Expression 8x-x^2 has a maximum value.
bi) Find maximum value

bii) State value of x for which the maximum value occurs



Also how can u solve the simultaneous equations:
4x^2-y^2=40
2x+y=10
Reply 1
First thing, multiply out that expression with ps and qs in it.

Then compare it with the 8x-x^2 one, and see what you see....
Reply 2
a) Expand p(xq)2p-{(x-q)}^2 and you should be able to see first what q is, and then using that to help you work out what p is

b) i) Using your factorised version, what must be in the brackets to make p(xq)2p-{(x-q)}^2 as big as possible? Use that to work out the maximum value
ii) You know numerically what's in the brackets, and also in terms of x and a constant (q)


For your simultaneous equations, rearrange the second to give you either x in terms of y or y in terms of x (I suggest y in terms of x, there's no division). Substitute that y into the first equation to get out a quadratic in x. Solve this to get two values of x. Substitute those two values of x into the second simultaneous equation to get y - do one x at a time and pair it with the y that you get out, usually in the form (x1y1)(x_1\,y_1), (x2y2)(x_2\,y_2)
rearrange 2x+y=10 into y= ... and then substitute into the other equation. When you find x, sub the value back into 2x+y=10 to get your answer.
Reply 4
ye but i dont know how to compare it can sum1 PLEASE type out all the working out
Reply 5
You can expand p(xq)2p-{(x-q)}^2 believe in yourself :p:

Write your expansion in the form
ax2+bx+cax^2 + bx + c and write
x2+8x+0{-x^2} +8x + 0 (your original equation) under it. You should be able to see what p and q are
Reply 6
Expansion:

p-(x-q)^2 = p-(x^2-2qx+q^2) = p-x^2+2qx-q^2

Compare:

p-x^2+2qx-q^2
8x-x^2

Rearrange:

-x^2+2qx-q^2+p --> a = -1, b = 2q, c = -q^2+p
-x^2+8x --> a = -1, b = 8, c = 0

form equations:

2q = 8
q = 4

-q^2+p = 0
-4^2+p = 0
-16+p = 0
p = 16
Reply 7
thnx 4 all ur replies i understand it now
Reply 8
Original post by *Princess*
thnx 4 all ur replies i understand it now


i have this same question and i know how to do part a) but i do not understand part b) i) find the maximum value of 8x-x^2 and ii) state the value of x for which this maximum occurs.

can you please help me :confused:
Reply 9
This is quite old thread!

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

16(x4)216-(x-4)^2

The maximum value of 8xx28x-x^2 is -16.

The maximum value of xx is 4.

The general formula for completing squares is:

a(x+h)2+ka(x+h)^2+k

aa is the maximum, but this needs to be a negative number. If the aa value in the equation if negative, leave it as it is, if it is positive, change it to negative.

hh is your maximum value for xx.

kk is the maximum value for yy.

So in this equation, the maximum coordinates are (4,0)(4, 0) because there is no kk value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how y=f(x+2)y=f(x+2) you move to the left rather than to the right and how y=f(2x)y=f(2x) is actually a stretch by scale factor 12\frac {1}{2}.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:
Wikipedia
Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/
(edited 13 years ago)
Reply 10
Original post by drewb
This is quite old thread!

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

16(x4)216-(x-4)^2

The maximum value of 8xx28x-x^2 is -16.

The maximum value of xx is 4.

The general formula for completing squares is:

a(x+h)2+ka(x+h)^2+k

aa is the maximum, but this needs to be a negative number. If the aa value in the equation if negative, leave it as it is, if it is positive, change it to negative.

hh is your maximum value for xx.

kk is the maximum value for yy.

So in this equation, the maximum coordinates are (4,0)(4, 0) because there is no kk value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how y=f(x+2)y=f(x+2) you move to the left rather than to the right and how y=f(2x)y=f(2x) is actually a stretch by scale factor 12\frac {1}{2}.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:
Wikipedia
Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/


Thank-you so much. I actually get it eventhough i thought it was very confusing at first. I also checked out the links that you attached and they were really helpful.

I tried the steps for a similar question like this and got it right

thanks again :smile: