[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

[Jam']

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.

[thorn0123]

(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

[Secreay]

thanks

[Secreay]

actually questions 2 is way to confusing could you give an example?

[Jam']

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 :/

[TDL]

(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.

[Jam']

(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

[TDL]

(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" ?

[Jam']

(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.

[Jam']

Also, the question would have given the symbol for the other variable to use if both of the sides weren't x.

[TDL]

meh, alright.

[TenOfThem]

There is nothing in the question that suggests an isosceles triangle

[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
Corrected now ... always check working

[TenOfThem]

(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

[TDL]

(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

[TenOfThem]

(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'

[TDL]

(Original post by TenOfThem)
Thanks

I knew I should have written it down

Principle stands tho'

just edited, my bad

[TenOfThem]

(Original post by TDL)
just edited, my bad

I have edited too

Thanks