Quadratic equation

**NyRychVantel** · October 11th 2008, 12:57 AM

The answers are p=3 q=2

I don't know the steps, pls help me

Thanks.

**TwistedOne151** · October 11th 2008, 01:41 AM

Note that $\frac{\alpha}{2}$ is a root of $2x^2+x+1$ . This means:
$2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\alpha^2=-\alpha-2$

Use this, along with $\alpha^4=(\alpha^2)^2$ , to get your result.

--Kevin C.

**NyRychVantel** · October 11th 2008, 01:56 AM

Originally Posted by TwistedOne151

Note that $\frac{\alpha}{2}$ is a root of $2x^2+x+1$ . This means:
$2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\alpha^2=-\alpha-2$

Use this, along with $\alpha^4=(\alpha^2)^2$ , to get your result.

--Kevin C.

I've tried that before,
$\begin{array}{l} \alpha ^2 = - 2 - \alpha \\ \alpha ^4 = \left( { - 2 - \alpha } \right)^2 = \alpha ^2 + 4\alpha + 4 \\ \alpha ^2 + 4\alpha + 4 = p\alpha + q \\ ????? \\ \end{array}$

**badgerigar** · October 11th 2008, 02:53 AM

Good job getting that far.
You still have an $\alpha^2$ . Use

again to get the answer

**NyRychVantel** · October 11th 2008, 03:17 AM

Originally Posted by badgerigar

Good job getting that far.
You still have an $\alpha^2$ . Use

again to get the answer

oh ya! stupid me, I didn't realize I can still sub $\alpha^2$ again!

Thanks.

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