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1. FP1 Complex
1)A and b , The sum of these 2 numbers is 10 and the product is 40

show that a must satisfy the equation a^2)-10a+40=0

2) A Right-Angled Triangle has area Acm^2 and perimeter Pcm. A side other than the hypotenuse has length has length xcm. Form a quadratic equation in x in each of the following cases:
a) a=6 p=12
b) a=3 p=8
c)a=30 p=30

This makes no sense to me at all could someone tell me what to do at all? thanks
2. Re: FP1 Complex
It's about the relationship between the coefficients of a polynomial and the roots of the polynomial:

For the equation ax^2+bx+c=0, with roots alpha and beta, the following is true:

Sum (alpha) = alpha + beta = -b/a = 10

Sum (alpha * beta) = alpha * beta = c/a = 40

Taking a to = 1:

-b = 10 => b = -10
c = 40

So:

alpha - 10 alpha + 40 = 0

Try and do question two with this idea.
Last edited by Jam'; 26-07-2012 at 02:14.
3. Re: FP1 Complex
(Original post by Secreay)
1)A and b , The sum of these 2 numbers is 10 and the product is 40

show that a must satisfy the equation a^2)-10a+40=0

2) A Right-Angled Triangle has area Acm^2 and perimeter Pcm. A side other than the hypotenuse has length has length xcm. Form a quadratic equation in x in each of the following cases:
a) a=6 p=12
b) a=3 p=8
c)a=30 p=30

This makes no sense to me at all could someone tell me what to do at all? thanks
for 1) just turn what they are saying into an equation i.e the sum of a and b is 10, so a+b = 10, the product is 40 so ab = 40, make a the subject and substitute it into the other equation and you'll get what they require.

2) Draw the triangle with the sides as they require and remember that a^2 + b^2 = c^2
4. Re: FP1 Complex
thanks
Last edited by Secreay; 26-07-2012 at 02:35.
5. Re: FP1 Complex
actually questions 2 is way to confusing could you give an example?
6. Re: FP1 Complex
I'm sure you can't have a right-angled isosceles triangle with area 6cm^2 and perimeter 12cm. I may be wrong but it doesn't seem to add up.

I mean the area, A would have to be as follows:

A= bh/2 = (x^2)/2
If A = 6, then (x^2)/2 = 6 => x = sqrt(12)

Our perimeter, P is (2+sqrt(2))x

But x = sqrt(12) is not a solution to (2+sqrt(2))x = 12

Maybe I've misunderstood the question. Sorry cant help there :/
Last edited by Jam'; 26-07-2012 at 05:55.
7. Re: FP1 Complex
(Original post by Secreay)
actually questions 2 is way to confusing could you give an example?
Take every information they give you and form equations out of it

so you know the area of a right angled triangle is (base*height*)/2 and the the perimeter is all the sides added so what you can do is let one of the sides called x (which they say is not the hypotenuse) let the hypotenuse be called h and the remaining side be y

so for the perimeter we get x + y + h = 12, rearrange this to get h = 12 - x - y
now for the area it is rearrange this to get now you know that we can find the hypotenuse of a right angled triangle by using Pythagoras theorem c^2 = a^2 + b^2, where c is the hypotenuse, so we just fill this in now sub in the equation of h you got earlier, as well as the equation of y to get now just expand the brackets and you will get a quadratic equation in x which should be Now this question is very vague, i'm not sure if they want you to find the hypotenuse but if they do it's the method i've shown if they don't then the equation will just be

You could of course juse let p=p and a=a and do the method I've should and you'll get a general equation where you can just sub p and a into.
8. Re: FP1 Complex
(Original post by TDL)
Take every information they give you and form equations out of it

so you know the area of a right angled triangle is (base*height*)/2 and the the perimeter is all the sides added so what you can do is let one of the sides called x (which they say is not the hypotenuse) let the hypotenuse be called h and the remaining side be y

so for the perimeter we get x + y + h = 12, rearrange this to get h = 12 - x - y
now for the area it is rearrange this to get now you know that we can find the hypotenuse of a right angled triangle by using Pythagoras theorem c^2 = a^2 + b^2, where c is the hypotenuse, so we just fill this in now sub in the equation of h you got earlier, as well as the equation of y to get now just expand the brackets and you will get a quadratic equation in x which should be Now this question is very vague, i'm not sure if they want you to find the hypotenuse but if they do it's the method i've shown if they don't then the equation will just be

You could of course juse let p=p and a=a and do the method I've should and you'll get a general equation where you can just sub p and a into.
x=y...if you read the question it says all sides other than the hypotenuse are x cm long. So, h = sqrt(2)x
9. Re: FP1 Complex
(Original post by Jam')
x=y...if you read the question it says all sides other than the hypotenuse are x cm long. So, h = sqrt(2)x
It says "A side other than the hypotenuse has length has length xcm" ?
10. Re: FP1 Complex
(Original post by TDL)
It says "A side other than the hypotenuse has length has length xcm" ?
Exactly - and there are two of them?

If I say, 'a person without hair is bald' that means every person that doesn't have hair is bald.
11. Re: FP1 Complex
Also, the question would have given the symbol for the other variable to use if both of the sides weren't x.
12. Re: FP1 Complex
meh, alright.
13. Re: FP1 Complex
There is nothing in the question that suggests an isosceles triangle
14. Re: FP1 Complex
Here is the solution for A=6 and P=12

Spoiler:
Show

Sides are a, b, c

Giving

x is not 0

Hmmm I missed post 7
Corrected now ... always check working
Last edited by TenOfThem; 26-07-2012 at 17:58. Reason: dimness
15. Re: FP1 Complex
(Original post by Jam')
Also, the question would have given the symbol for the other variable to use if both of the sides weren't x.
this is simply not true

there is enough information to solve this

the assumption that it is an isosceles triangle is false
16. Re: FP1 Complex
(Original post by TenOfThem)
Here is the solution for A=6 and P=12

Spoiler:
Show

Sides are a, b, c

Giving

x is not 0

Hmmm I missed post 7
My result is slightly different
you forgot to add 24x^2 to 144x^2
Last edited by TDL; 26-07-2012 at 17:56.
17. Re: FP1 Complex
(Original post by TDL)
you expanded wrong: (12x-x^2-12)^2 = 144x^2 -12x^3 - 144x - 12x^2 + x^4 + 12x^2 -144x +12x^2 + 144

which gives
168x^2 - 24x^3 - 288x + x^4 + 144
Thanks

I knew I should have written it down

Principle stands tho'
18. Re: FP1 Complex
(Original post by TenOfThem)
Thanks

I knew I should have written it down

Principle stands tho'
just edited, my bad
19. Re: FP1 Complex
(Original post by TDL)
just edited, my bad
I have edited too

Thanks
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