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Roots of Polynomial Equations

The equation x2+2x+5=0x^{2} + 2x + 5 = 0 has roots α\alpha and β\beta. Find the equations which have the following roots.

α2\alpha^2 , β2\beta^2

I've tried letting the roots equal u. Rearrange for α\alpha and β\beta and substitute them into the equation. But it hasn't worked, any ideas how to do this?


α+2β,β+2α\alpha + 2\beta , \beta + 2\alpha

On this one again, I've let roots equal u. Then by using simultaneous equation I eliminated β\beta and substituted my α\alpha into the quadratic. Again I get the wrong. Is this method not suitable for this kind of question?
(edited 13 years ago)
You can do it the way you said (but it is not the most efficient way).

u=α2u = \alpha^2

α=±u\alpha = \pm \sqrt u

u+2(±u)+5=0u + 2 (\pm \sqrt u) + 5 = 0

Rearrange to make ±u\pm \sqrt u the subject and square.

It is easier to use the identity α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta

You can use a u substitution for the second one too.

u=α+2βu = \alpha + 2 \beta

You know α+β=−2\alpha + \beta = -2 so u=−2+βu = -2 + \beta and β=u+2\beta = u + 2
(edited 13 years ago)
Reply 2
Avatar for Freerider101
Freerider101
OP
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Original post by Mr M
You can do it the way you said (but it is not the most efficient way).

u=α2u = \alpha^2

α=±u\alpha = \pm \sqrt u

u+2(±u)+5=0u + 2 (\pm \sqrt u) + 5 = 0

Rearrange to make ±u\pm \sqrt u the subject and square.

It is easier to use the identity α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta

You can use a u substitution for the second one too.

u=α+2βu = \alpha + 2 \beta

You know α+β=−2\alpha + \beta = -2 so u=−2+βu = -2 + \beta and β=u+2\beta = u + 2


Thank you. For the first part isn't the answer just u+2(±u)+5=0u + 2 (\pm \sqrt u) + 5 = 0 squared as it wanted the equation or is this why I'm not getting the same answer as the textbook?

And for the second part, I like your method. I'm just wondering why my original method didn't work.

u=α+2β u = \alpha + 2 \beta can be rearranged as α=u−2β \alpha = u - 2\beta
and
u=β+2α u = \beta + 2\alpha can be rearranged as β=u−2α \beta = u - 2\alpha

I can substitute 2nd equation into 1st to get

α=u−2(u−2α) \alpha = u - 2(u -2\alpha)

α=−u+4α \alpha = -u + 4\alpha

3α=u 3\alpha = u

α=u3 \alpha = \frac{u}{3}

And this value for α \alpha is different to yours, so I must have gone wrong with my method some how?
Original post by Freerider101
Thank you. For the first part isn't the answer just u+2(±u)+5=0u + 2 (\pm \sqrt u) + 5 = 0 squared as it wanted the equation or is this why I'm not getting the same answer as the textbook?

And for the second part, I like your method. I'm just wondering why my original method didn't work.

u=α+2β u = \alpha + 2 \beta can be rearranged as α=u−2β \alpha = u - 2\beta
and
u=β+2α u = \beta + 2\alpha can be rearranged as β=u−2α \beta = u - 2\alpha

I can substitute 2nd equation into 1st to get

α=u−2(u−2α) \alpha = u - 2(u -2\alpha)

α=−u+4α \alpha = -u + 4\alpha

3α=u 3\alpha = u

α=u3 \alpha = \frac{u}{3}

And this value for α \alpha is different to yours, so I must have gone wrong with my method some how?


On the first thing you need to make the ±\pm bit the subject so that it becomes positive when you square both sides.

On your second point, you are making the incorrect assumption that the roots of the new equation are equal (you have set them both equal to u).

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