-mel-
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#1
Report Thread starter 11 years ago
#1
Can somebody please tell me if I have the eqn 2x^2 + 4x + 3 =0

(Say alpha is a and beta is b)

I know how to work out (a^4)(b^4)

But how do you work out a^4 + b^4?

I know that a^2 + b^2 => (a+b)^2 - 2ab

But can somebody tell me what the solution is to the power of 4. The exam is on Monday and this is something I need to know!

Thanks!
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2^1/2
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#2
Report 11 years ago
#2
a and b are the roots of the equation.

you need to use the fact that, a^2 + b^2 = (a + b)^2 - 2ab

a =  \alpha , b =  \beta
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Krazof
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#3
Report 11 years ago
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Try expanding (a+b)^4 and see if that helps you.
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-mel-
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#4
Report Thread starter 11 years ago
#4
I've tried but I just keep going round and round in circles :|
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nota bene
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#5
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(a^2+b^2)^2=((a+b)^2)^2-4ab(a+b)^2+4a^2b^2 etc.?
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username167718
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#6
Report 11 years ago
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a^4 + b^4



= (a+b)^4 - 4a^3b - 6a^2b^2 - 4ab^3



= (a+b)^4 - 2(ab)(2a^2 + 3ab + 2b^2)



= (a+b)^4 - 2(ab)(2(a+b)^2 - (ab))

Notice how I put all the (a+b) and (ab) in brackets? I did that to highlight the main point here:
The key to any of these questions is to rearrange the equation in terms of the sum of the roots, and the product of the roots (and with more complex problems, such as cubics and quartics, you also use the sum of the roots in pairs etc.)
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