Roots of polynomial equations 3

[Erfan1]

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]

I would begin with:

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

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:

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

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

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

 

[anemone]

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)