Sum and product of roots, this is so confusing!

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February 5th 2011, 02:31 AM
PythagorasNeophyte

Sum and product of roots, this is so confusing!

Question:
The roots of the equation $3x^2 - 3kx + k - 6 = 0$ are $\alpha$ and $\beta$ . If $\alpha^2 + \beta^2$ = $\dfrac{20}{3}$ , find the possible values of $k$ .

My workings:
$\alpha + \beta =$ $\dfrac{-b}{a}$ $= \dfrac{3k}{3} = k$

$\alpha\beta =$ $\dfrac{c}{a}$ $= \dfrac{k-6}{3}$

Can anyone facilitate in solving this problem?
February 5th 2011, 02:39 AM
DrSteve

Use the fact that the product of the roots is $\frac{k-6}{3}$ and the sum of the roots is $k$ (which you showed), together with $(\alpha +\beta )^2 = \alpha^2 + 2\alpha \beta + \beta^2$ .
February 5th 2011, 02:47 AM
PythagorasNeophyte

Hello DrSteve:

$(\alpha +\beta )^2 = \alpha^2 + 2\alpha \beta + \beta^2$

$k^2 = \alpha^2 + \beta^2 + 2\alpha\beta$

$k^2 = (\alpha+\beta)(\alpha+\beta) + 2(\dfrac{k-6}{3})$

$k^2 = k^2 + 2(\dfrac{k-6}{3})$

Is this correct?
February 5th 2011, 02:56 AM
CaptainBlack

Quote:

Originally Posted by PythagorasNeophyte

Hello DrSteve:

$(\alpha +\beta )^2 = \alpha^2 + 2\alpha \beta + \beta^2$

$k^2 = \alpha^2 + \beta^2 + 2\alpha\beta$

$k^2 = (\alpha+\beta)(\alpha+\beta) + 2(\dfrac{k-6}{3})$

$k^2 = k^2 + 2(\dfrac{k-6}{3})$

Is this correct?

You can check yourself: from that last equation solve for the numerical value of k, substitute back into the original equation and solve to find the roots and check that the sum of the squares of the roots is 20/3.

And the answer is no, $\alpha^2+\beta^2 \ne (\alpha+\beta)(\alpha+\beta)$ , you are told:

$\alpha^2+\beta^2 = \frac{20}{3}$

CB
February 5th 2011, 03:02 AM
DrSteve

Quote:

Originally Posted by PythagorasNeophyte

Hello DrSteve:

$(\alpha +\beta )^2 = \alpha^2 + 2\alpha \beta + \beta^2$

$k^2 = \alpha^2 + \beta^2 + 2\alpha\beta$

$k^2 = (\alpha+\beta)(\alpha+\beta) + 2(\dfrac{k-6}{3})$

$k^2 = k^2 + 2(\dfrac{k-6}{3})$

Is this correct?

You have an error. From your second step, just replace $\alpha^2 + \beta^2$ by $\frac{20}{3}$

$k^2 = \frac{20}{3} + 2(\dfrac{k-6}{3})$

Edit: Just noticed that CaptainBlack already told you this - now you've heard it twice. (Nod)
February 5th 2011, 06:23 AM
CaptainBlack

Quote:

Originally Posted by DrSteve

You have an error. From your second step, just replace $\alpha^2 + \beta^2$ by $\frac{20}{3}$

$k^2 = \frac{20}{3} + 2(\dfrac{k-6}{3})$

Edit: Just noticed that CaptainBlack already told you this - now you've heard it twice. (Nod)

Once more and we all know its true :)

CB