1. ## Roots of polynomials

Hi, sorry if this is in the wrong forums not really sure but here it goes..

The quadratic equation $\displaystyle 2x^2+4x+3=0$ has roots $\displaystyle \alpha$ and $\displaystyle \beta$ .

Find the value $\displaystyle \alpha^4+\beta^4$

Ive managed to expand to $\displaystyle \alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$

but im not sure how to put in terms of $\displaystyle \alpha\beta$ and $\displaystyle \alpha+\beta$ to work out an answer.

Also the previous part of the question asks to show $\displaystyle \alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?

Help really appreciated. Thankyou !

2. Hello, NathanBUK!

The quadratic equation $\displaystyle 2x^2+4x+3\:=\:0$ has roots $\displaystyle \alpha$ and $\displaystyle \beta$ .

Find the value of: $\displaystyle \alpha^4+\beta^4$

Also the previous part of the question asks to show $\displaystyle \alpha^2+\beta^2=1$
. . which I managed to prove. . . . . Good!

Is there anyway I can use this to answer the question? . . . . Yes!

We know that: .$\displaystyle \begin{Bmatrix}\alpha + \beta &=& -2 \\ \alpha\beta &=& \frac{3}{2} \end{Bmatrix}$

We have: .$\displaystyle \alpha^2 + \beta^2 \;=\;1$

Square: .$\displaystyle \left(\alpha^2 + \beta^2\right)^2 \;=\;(1)^2$

. . . . $\displaystyle \alpha^4 + 2\alpha^2\beta^2 + \beta^4 \;=\;1$

. . . $\displaystyle \alpha^4 + \beta^4 + 2\!\cdot\!\!\!\underbrace{(\alpha\beta)^2}_{\text{ This is }(\frac{3}{2})^2} \;=\;1$

. . . . . $\displaystyle \alpha^4 + \beta^4 + \frac{9}{2} \;=\;1$

. . . . . . . $\displaystyle \alpha^4 + \beta^4 \;=\;-\frac{7}{2}$

3. Ohh thats makes soo much sense now!! Thankyou so much!!

4. Originally Posted by NathanBUK
Hi, sorry if this is in the wrong forums not really sure but here it goes..

The quadratic equation $\displaystyle 2x^2+4x+3=0$ has roots $\displaystyle \alpha$ and $\displaystyle \beta$ .

Find the value $\displaystyle \alpha^4+\beta^4$

Ive managed to expand to $\displaystyle \alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$

but im not sure how to put in terms of $\displaystyle \alpha\beta$ and $\displaystyle \alpha+\beta$ to work out an answer.

Also the previous part of the question asks to show $\displaystyle \alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?

Help really appreciated. Thankyou !
Alternative solution (same one in disguise realy):

$\displaystyle \alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha \beta)^2$

CB