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

phys981

10

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

Trangulor

6

a) Expand $p-{(x-q)}^2$ 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 $p-{(x-q)}^2$ 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 $(x_1\,y_1)$ , $(x_2\,y_2)$

Reply 3

sugarwhirl89

1

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

*Princess*

OP

ye but i dont know how to compare it can sum1 PLEASE type out all the working out

Reply 5

Trangulor

6

You can expand $p-{(x-q)}^2$ believe in yourself :p:

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

Reply 6

dezlee

7

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

*Princess*

OP

thnx 4 all ur replies i understand it now

Reply 8

hyy

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

kanojyoxx

12

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:

$16-(x-4)^2$

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

The maximum value of $x$ is 4.

The general formula for completing squares is:

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

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

$h$ is your maximum value for $x$ .

$k$ is the maximum value for $y$ .

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ 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 $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$ .

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 14 years ago)

Reply 10

hyy

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:

$16-(x-4)^2$

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

The maximum value of $x$ is 4.

The general formula for completing squares is:

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

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

$h$ is your maximum value for $x$ .

$k$ is the maximum value for $y$ .

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ 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 $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$ .

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:

Completing the square

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