Question

HELP MEDAL GIVEN!!!

Question Down Below I
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Answer 1

The Quadratic Equation : \[2x^2 + 4x + 3 = 0\]

I have done part a) and b)

so part c question is :

find the value of : \[\alpha^4 + \beta^4\]

Answer 2

@Nnesha @nincompoop @paki

Answer 3

@Luigi0210 @campbell_st

Answer 4

well

\[\alpha^4 + \beta^4= (\alpha^2 + \beta^2)^2 - 2 \alpha^2 \beta^2\]

so that is probably a start

Answer 5

so apply the same reasoning to deal with

\[(\alpha^2 + \beta^2)^2\]

so that it is in terms of

\[\alpha + \beta\]

Answer 6

ok... so

\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\]

so calculate this and then substitute it

Answer 7

and in the original equation

\[2\alpha^2 beat^2 = 2(\alpha \beta)^2\]

Answer 8

so for that I got the answer = 1

Answer 9

well that makes sense

so you have

\[(1)^2 - 2(\alpha \beta)^2\]

Answer 10

a medal and i will help

Answer 11

so, that equals = -7/2

Answer 12

that's it