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Roots of polynomial equations 3

Erfan

New member
Jul 19, 2013
9
Obtain the sum of the squares of the roots of the equation x^4 + 3x^3 + 5x^2 + 12x + 4 = 0 .
Deduce that this equation does not have more than 2 real roots .
Show that , in fact , the equation has exactly 2 real roots in the interval -3 < x < 0 .
Denoting these roots α and β , and the other 2 roots by γand δ , show that :
modulus of γ = modulus of δ = 2/(radical(αβ) .
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would begin with:

\(\displaystyle f(x)=x^4+3x^3+5x^2+12x+4\)

Obviously, the real roots must be negative, and the rational roots theorem shows us that none are rational.

So, let's see if this quartic can be factored:

\(\displaystyle f(x)=x^2\left(x^2+4 \right)+3x\left(x^2+4 \right)+\left(x^2+4 \right)\)

\(\displaystyle f(x)=\left(x^2+3x+1 \right)\left(x^2+4 \right)\)

Now you may obtain the roots explicitly, simplifying matters greatly.
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Obtain the sum of the squares of the roots of the equation x^4 + 3x^3 + 5x^2 + 12x + 4 = 0 .
Deduce that this equation does not have more than 2 real roots .
Show that , in fact , the equation has exactly 2 real roots in the interval -3 < x < 0 .
Denoting these roots α and β , and the other 2 roots by γand δ , show that :
modulus of γ = modulus of δ = 2/(radical(αβ) .
Let $\alpha, \beta, \gamma$ and $\delta$ be the roots of the equation $x^4+3x^3+5x^2+12x+4=0$.

If we're allowed to use Newton's Identities to solve the first part of the problem, we see that

$(\alpha+ \beta+ \gamma+\delta)(1)+(1)(3)=0\;\;\rightarrow\;\alpha+ \beta+ \gamma+\delta=-3$

$(\alpha^2+ \beta^2+ \gamma^2+\delta^2)(1)+(3)(\alpha+ \beta+ \gamma+\delta)+(2)(5)=0\;\;\rightarrow\;\alpha^2+ \beta^2+ \gamma^2+\delta^2=-1$

Descartes' Rule of Signs tell us [$f(x)=x^4+3x^3+5x^2+12x+4$ and $f(-x)=x^4-3x^3+5x^2-12x+4$] there are 4, 2 or 0 negative solutions and the function $f(x)$ has no positive solutions.

Since $\alpha^2+ \beta^2+ \gamma^2+\delta^2=-1$, it can't be the case where all 4 of the roots are real negative roots. So, we can deduce now that the given equation does not have more than 2 real roots .

Notice that $f(-3)=13$, $f(-2)=-8$, so we know that there is a root in the interval $-3<x<-2$ by the Intermediate Value Theorem.

Also $f(-1)=-5$, $f(0)=4$, the other negative root in the interval $-1<x<-0$.

Combining these two results shows that the equation has exactly 2 real roots in the interval $-3<x<0$ .

Let both of the imaginary roots of the equation $\gamma$ and $\delta$ be expressed in the form $\gamma=a+bi$ and $\delta=a-bi$, we're asked to prove $|\gamma|=|\delta|=\frac{2}{\sqrt{\alpha\beta}}$ is true, or equivalently, $\sqrt{a^2+b^2}=\frac{2}{\sqrt{\alpha\beta}}$ is true.

Vieta's formula tells us the product of the 4 roots $\alpha, \beta, \gamma$ and $\delta$ is 4.

$\alpha\beta\gamma\delta=4$

$(a+bi)(a-bi)\alpha\beta=4$

$(a^2+b^2)\alpha\beta=4$

$a^2+b^2=\frac{4}{\alpha\beta}$

$\therefore \sqrt{a^2+b^2}=\frac{2}{\sqrt{\alpha\beta}}$(QED)
 
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