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# alpha and beta quadratic equation roots watch

1. Hi I'm trying to solve the roots of x^2+5x+3=0 whose roots are 2α+ß and α+2ß.

I've already solved the sum which is -15 but I'm having trouble with the product.

I multiplied the roots to give me 2α^2 +5αß + 2ß^2 but now I don't know what to do for the α^2 and ß^2. how do I get the values for these?

btw I know that αß=3 and α+ß= -5

thanks
2. (Original post by HelloGoodbye)
Hi I'm trying to solve the roots of x^2+5x+3=0 whose roots are 2α+ß and α+2ß.

I've already solved the sum which is -15 but I'm having trouble with the product.

I multiplied the roots to give me 2α^2 +5αß + 2ß^2 but now I don't know what to do for the α^2 and ß^2. how do I get the values for these?

btw I know that αß=3 and α+ß= -5

thanks
Are you trying to find the new equation

You need the standard result for (a+b)^2
3. (Original post by TenOfThem)
Are you trying to find the new equation

You need the standard result for (a+b)^2
yes I am and so should I rearrange 2α^2 + 5αß +2ß^2 to 5αß + (α+ß)^2?

thanks!
4. (Original post by HelloGoodbye)
yes I am and so should I rearrange 2α^2 + 5αß +2ß^2 to 5αß + (α+ß)^2?

thanks!
No

They are not the same

You should know that (a+b)^2 = a^2+b^2+2ab
5. You could try expanding (x - (2a+b)) (x - (a+2b)) and comparing the results.

Spoiler:
Show
x2 - ax - 2bx - 2ax -bx + (2a+b)(a+2b)
= x2 - 3ax -3bx + 2a2 + 5ab + 2b2

So by comparing this to x2+5x+3 we get:

-3ax-3bx = 5x
a+b=-5/3

2a2 + 5ab + 2b2 = 3

Substitute or simultaneous to work out a and b

edit: looking at your given answers I did something wrong here but it's late, I promise the method works :P
6. (Original post by TenOfThem)
No

They are not the same

You should know that (a+b)^2 = a^2+b^2+2ab
so does that mean I should substitute (α+β)^2 - 2αβ into the equation? I'm so confused!!
7. (Original post by HelloGoodbye)
so does that mean I should substitute (α+β)^2 - 2αβ into the equation? I'm so confused!!
Yes

I don't know why you are confused

This is a standard method ... It will be in your textbook and I would expect it to have been discussed in class

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