Find value of expression involving roots of a quadratic

**perrification** · September 27th 2011, 12:01 PM

The roots of the equation x^2 - 2x + 7=0 are alpha and beta. Without finding the value of alpha and beta explicitly, determine 2B^3 - 3a^2 + 6 (a being alpha and B being beta)

i post it here as i feel that its a complex number question but i found this question from grade 9 where complex numbers have yet to be introduced (singapore)

really desperate to solve this question. the answer seem to be a clear cut integer instead of multiple answers in complex form.

by the way, this has to do with sum and product of roots

**ModusPonens** · September 27th 2011, 12:26 PM

I haven't solved the whole problem by hand, but here's the idea: you aply the division algorithm to the polynomials 2x^3 with divisor x^2-2x+7 and the polynomial -3x^2 with the same divisor as the anterior. Next you substitute the B and a in the quotient and it is equal to 0. you will then have only operations on numbers.

**Soroban** · September 27th 2011, 04:30 PM

Hello, perrification!

Well, I've made some progress, but . . .

$\text{The roots of the equation }x^2 - 2x + 7\:=\:0\,\text{ are }\alpha\text{ and }\beta.$
$\text{Without finding the value of }\alpha\text{ and }\beta\text{ explicitly, determine: }\:2\beta^3 - 3\alpha^2 + 6$

I post it here as i feel that its a complex number question, but i found this question
from grade 9 where complex numbers have yet to be introduced (Singapore).

Really desperate to solve this question. the answer seem to be a clear-cut integer
instead of multiple answers in complex form.

By the way, this has to do with sum and product of roots.

Since $\alpha$ and $\beta$ are roots of the first equation: . $\begin{Bmatrix}\alpha + \beta &=& 2 & [1]\\ \alpha\!\cdot\!\beta &=& 7 & [2]\end{Bmatrix}$

Square [1]: . $(\alpha + \beta)^2 \:=\:2^2 \quad\Rightarrow\quad \alpha^2 +2(\alpha\beta) + \beta^2\:=\:4$

Substitute [2]: . $\alpha^2 + 2(7) + \beta^2 \:=\:4 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:\text{-}10$

It seems that we are dealing with complex numbers.

We have: . $\alpha^3 + \beta^3 \;=\;(\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \;=\;(2)(\text{-}10 - 7) \;=\;-34$

So we have: . $\begin{Bmatrix} \alpha\cdot\beta &=& 7 \\ \alpha + \beta &=& 2 \\ \alpha^2 + \beta^2 &=& \text{-}10 \\ \alpha^3 + \beta^3 &=& \text{-}34 \end{Bmatrix} \;\;\hdots\:\text{ all real numbers!}$

Does this help?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I did it "the long way" and got a surprising answer.

The roots of the quadratic are: . $x \:=\:1 \pm i\sqrt{6}$

Let: . $\begin{Bmatrix} \alpha &=& 1 + i\sqrt{6} \\ \beta &=& 1 - i\sqrt{6} \end{Bmatrix}$

Then: . $\begin{Bmatrix}\alpha^2 &=& -5 + 2i\sqrt{6} \\ \beta^3 &=& -17 + 3i\sqrt{6} \end{Bmatrix}$

Therefore: . $2\beta^3-3\alpha^2 + 6 \;\;=\;\;2(-17 + 3i\sqrt{6}) - 3(-5 + 2i\sqrt{6}) + 6$

. . . . . . . . . . . $=\;\; -34 + 6i\sqrt{6} + 15 - 6y\sqrt{6} + 6 \;\;=\;\;\boxed{-13} \quad ha!$

**perrification** · September 27th 2011, 09:29 PM

but if we were restricted with not using complex number it is most likely more on the manipulation of the alpha and betas.

**ModusPonens** · September 28th 2011, 05:18 AM

Have you understood my answer or do you want me to explain it more explicitly?

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