2
$\begingroup$

I was given this problem at school to look at home as a challenge, after spending a good 2 hours on this I can't seem to get further than the last part of the equation. I'd love to see the way to get through 2) before tomorrow's lesson as a head start.

So the problem is as follows:

1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$

i. Find roots $$\alpha + \beta$$

ii. Find roots $$\alpha\beta$$

2) Find an Equation with integer coefficients who's roots are:

$$2\alpha^4+\frac{1}{\beta^2}$$$$2\beta^4+\frac{1}{\alpha^2}$$

I'm completely puzzled on the second part of the question and I've tried following the method I was taught. Sorry if formatting is a bit off, first time posting here :)

Thanks in advance for any help!

$\endgroup$
3
  • 2
    $\begingroup$ Can you describe the method you was taught, so we can provide a hint on how to use that, instead of giving you hints you might not understand? $\endgroup$ Sep 16, 2015 at 19:36
  • $\begingroup$ If a quadratic equation has roots $r_1$ and $r_2$, it is of the form $a(x-r_1)(x-r_2)$. For part 1, let $r_1$ and $r_2$ be $\alpha$ and $\beta$ respectively, multiply out the above formula, and match coefficients. For part 2, let $r_1$ and $r_2$ be the expressions you've given. $\endgroup$ Sep 16, 2015 at 19:44
  • 1
    $\begingroup$ We were taught that an equation with 2 solutions can be displayed in a form of $$a^2-(a+b)x + ab = 0$$. Following this, we'd put in the roots given. I got to this point $$x^2 - ((2\alpha^4+\frac{1}{\beta^2})+(2\beta^4+\frac{1}{\alpha^2}))x +(2\alpha^4+\frac{1}{\beta^2})(2\beta^4+\frac{1}{\alpha^2})$$ $\endgroup$ Sep 16, 2015 at 19:53

3 Answers 3

1
$\begingroup$

Given $\displaystyle \alpha,\beta$ are the roots of $2x^2+8x+4=0.$

So$\displaystyle \alpha+\beta = -\frac{8}{2}=-4$ and $\displaystyle \alpha\cdot \beta = \frac{1}{2}.$

Now for Second part, Using $\bullet\; \bf{x^2-(sum \; of \; roots)x+(product\; of \; roots) =0}$

So here $\displaystyle \bf{sum\; of \; roots } = 2\alpha^4+\frac{1}{\beta^2}+2\beta^4+\frac{1}{\alpha^2} = 2\left[\alpha^4+\beta^4\right]+\frac{1}{\alpha^2}+\frac{1}{\beta^2}$

So we get $\displaystyle = 2\left[(\alpha^2+\beta^2)^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2}$

$\displaystyle = 2\left[\left\{(\alpha+\beta)^2-2\alpha\cdot \beta\right\}^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2} = $

and $\displaystyle \bf{product\; of roots} = \left(2\alpha^4+\frac{1}{\beta^2}\right)\times \left(2\beta^4+\frac{1}{\alpha^2}\right)$

$\displaystyle = 4(\alpha\cdot \beta)^4+2\left[(\alpha+\beta)^2-2\alpha\cdot \beta\right]+\frac{1}{(\alpha\cdot \beta)^2}=$

$\endgroup$
1
  • $\begingroup$ You meant $2x^2 + 8x + \color{red}{1} = 0$. $\endgroup$ Sep 26, 2015 at 0:28
0
$\begingroup$

1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$

i. $$\alpha + \beta=\frac{-b}{a}=\frac{-8}{2}=-4$$

ii. $$\alpha\beta=\frac{c}{a}=\frac{1}{2}$$ For remain just find roots and use this fact that if $x_1+x_2=s$ and $x_1 x_2=p$ equation will be $x^2 -sx+p=0$.

$\endgroup$
7
  • $\begingroup$ Thanks for the comment, I got this far...Its the second part i'm stuck on :/ $\endgroup$ Sep 16, 2015 at 19:42
  • $\begingroup$ I now will complete it $\endgroup$
    – R.N
    Sep 16, 2015 at 19:43
  • $\begingroup$ Isn't the sum of roots -b/a not -b/2a? $\endgroup$ Sep 16, 2015 at 19:44
  • $\begingroup$ $‎\frac {-b+‎\sqrt{b^2 -4ac}‎}{2a}+\frac {-b-\sqrt{b^2 -4ac}‎}{2a}=\dfrac{-b}{2a}$ $\endgroup$
    – R.N
    Sep 16, 2015 at 19:49
  • $\begingroup$ @RaziehNoori Double-check that. I think you'd get $\frac{-2b}{2a}=-\frac b{2a}$. $\endgroup$ Sep 16, 2015 at 20:10
0
$\begingroup$

Hint:

You must calculate $S=\alpha^4+\dfrac1{\beta^2}+\beta^4+\dfrac1{\alpha^2}$ and $P=\Bigl(\alpha^4+\dfrac1{\beta^2}\Bigr)\Bigl(\beta^4+\dfrac1{\alpha^2}\Bigr)$. They will be the roots of the quadratic equation $\;x^2-Sx+P=0$.

Now any symmetric rational function of $\alpha$ and $\beta$, by a theorem of Newton, can be expressed as a rational function of the elementary symmetric functions $s=\alpha+\beta$ and $p=\alpha\beta$: \begin{align*} \alpha^2+\beta^2&=s^2-2p,\enspace\text{hence}\quad\frac1{\alpha^2}+\frac1{\beta^2}=\frac{s^2-2p}{p^2},\\ \alpha^4+\beta^4&=\bigl(\alpha^2+\beta^2)^2-2\alpha^2\beta^2=(s^2-2p)^2-2p^2=s^2-4ps^2+2p^2 \end{align*} whence $S$.

Similarly: $$P=\alpha^4\beta^4+\alpha^2+\beta^2+\frac1{\alpha^2\beta^2}=p^4+s^2-2p+\frac1{p^2}.$$

$\endgroup$
5
  • $\begingroup$ Appreciate the answer, but in the way you have described, I don't quite understand what you mean...I've not been taught this methodology yet :P Thanks anyhow! $\endgroup$ Sep 16, 2015 at 20:21
  • $\begingroup$ It just uses high school formulae; I made computations with variables because they're easier to follow, but in the present case you just have to replace $s$ and $p$ with their values ($-4$ and $\dfrac12$ respectively). $\endgroup$
    – Bernard
    Sep 16, 2015 at 20:34
  • $\begingroup$ You never explicitly mentioned that $p = \alpha\beta$. Also, check your final formula for $\alpha^4 + \beta^4$. You did not expand $(s^2 - 2p)^2$ correctly. $\endgroup$ Sep 26, 2015 at 0:32
  • $\begingroup$ Thanks for pointing the typos! Corrected now. As for $p=\alpha\beta$, it must have indeed been hard to follow the computation for beginners! $\endgroup$
    – Bernard
    Sep 26, 2015 at 0:39
  • $\begingroup$ For $\alpha^4 + \beta^4$, I am getting $s^\color{red}{4} - 4ps^2 + 2p^2$. $\endgroup$ Sep 26, 2015 at 1:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .