MHB Roots of polynomial equations 1

[Erfan1]

Find the sum of the squares of the roots of the equation x^3 + x + 12 = 0 and deduce that only one of the roots is real .
The real root of the equation is denoted by alpha . Prove that -3< alpha < -2 , and hence prove that the modulus of each of the other roots lies between 2 and root 6 . I found the sum of the squares of the roots which is -2 but couldn't solve the other parts ! Thanks in advance !

 

[MarkFL]

Re: Roots of polynomial equations

If we let the three roots be $$\alpha,\,z_1,\,z_2$$ and write:

$$(\alpha+z_1+z_2)^2=\alpha^2+z_1^2+z_2^2+2(\alpha z_1+\alpha z_2+z_1z_2)$$

$$\alpha^2+z_1^2+z_2^2=(\alpha+z_1+z_2)^2-2(\alpha z_1+\alpha z_2+z_1z_2)$$

By Vieta, we have:

$$\alpha+z_1+z_2=-\frac{0}{1}=0$$

$$\alpha z_1+\alpha z_2+z_1z_2)=\frac{1}{1}=1$$

Hence:

$$\alpha^2+z_1^2+z_2^2=0^2-2(1)=-2$$

Because the sum of the squares of the 3 roots is negative, we know the roots are not all real, and by the conjugate roots theorem, we know there must be two complex conjugate roots.

To find the two integers between which $\alpha$ must lie, consider writing the cubic in the form:

$$x^3+x=-12$$

For the moduli of the complex roots, consider:

$$z_1=r\left(\cos(\theta)+i\sin(\theta) \right),\,z_2=r\left(\cos(\theta)-i\sin(\theta) \right)$$

Now, we may state (using Vieta again):

$$\alpha z_1z_2=-\frac{12}{1}=-12$$

Putting this together with the polar form of the complex roots, what do you find?

 

[Erfan1]

Re: Roots of polynomial equations

If we let the three roots be $$\alpha,\,z_1,\,z_2$$ and write:

$$(\alpha+z_1+z_2)^2=\alpha^2+z_1^2+z_2^2+2(\alpha z_1+\alpha z_2+z_1z_2)$$

$$\alpha^2+z_1^2+z_2^2=(\alpha+z_1+z_2)^2-2(\alpha z_1+\alpha z_2+z_1z_2)$$

By Vieta, we have:

$$\alpha+z_1+z_2=-\frac{0}{1}=0$$

$$\alpha z_1+\alpha z_2+z_1z_2)=\frac{1}{1}=1$$

Hence:

$$\alpha^2+z_1^2+z_2^2=0^2-2(1)=-2$$

Because the sum of the squares of the 3 roots is negative, we know the roots are not all real, and by the conjugate roots theorem, we know there must be two complex conjugate roots.

To find the two integers between which $\alpha$ must lie, consider writing the cubic in the form:

$$x^3+x=-12$$

For the moduli of the complex roots, consider:

$$z_1=r\left(\cos(\theta)+i\sin(\theta) \right),\,z_2=r\left(\cos(\theta)-i\sin(\theta) \right)$$

Now, we may state (using Vieta again):

$$\alpha z_1z_2=-\frac{12}{1}=-12$$

Putting this together with the polar form of the complex roots, what do you find?

I'm not really good at complex number .

 

[MarkFL]

Re: Roots of polynomial equations

What is:

$$z_1z_2=r\left(\cos(\theta)+i\sin(\theta) \right)\cdot r\left(\cos(\theta)-i\sin(\theta) \right)$$ ?

Or, to avoid trigonometry, let:

$$z_1=a+bi$$

$$z_2=a-bi$$

Thus:

$$z_1z_2=(a+bi)(a-bi)$$

In both cases, recall:

$$i^2=-1$$

and for the polar form, recall:

$$\sin^2(\theta)+\cos^2(\theta)=1$$

 

[zzephod]

Re: Roots of polynomial equations

Find the sum of the squares of the roots of the equation x^3 + x + 12 = 0 and deduce that only one of the roots is real

Descartes' rule of signs tells you that there are no positive roots and 1 negative root. Hence there is exactly 1 real root and hence the other two are complex.

.

 

[zzephod]

Re: Roots of polynomial equations

Find the sum of the squares of the roots of the equation x^3 + x + 12 = 0 and deduce that only one of the roots is real .
The real root of the equation is denoted by alpha . Prove that -3< alpha < -2 ,

$$f(x)=x^3 + x + 12$$ is -ve at $$x=-3$$ and +ve at $$x=-2$$ hence $$f(x)$$ has a root in the interval $$(-3,-2).$$

.

 

[I like Serena]

Re: Roots of polynomial equations

I'm not really good at complex number .

A third order polynomial has at least one real root, since it tends to different infinities in plus and minus directions.
If it has 2 real roots, one of those roots must be a duplicate root, meaning there are really 3 real roots.
So either we have 1 real root and 2 imaginary roots, or we have 3 real roots (which may contain duplications).

With MarkFL's $$\alpha^2+z_1^2+z_2^2=0^2-2(1)=-2$$ it follows that there must be imaginary roots, since the squares of real numbers are always at least 0.
In other words, there is 1 real root and 2 imaginary roots.

Since there is only one real root, simply filling in x=-2 respectively x=-3 yields that they are on opposite sides of the x-axis.
Therefore the 1 real root must be between -2 and -3.

 

[Deveno]

Re: Roots of polynomial equations

A third order polynomial has at least one real root, since it tends to different infinities in plus and minus directions.
If it has 2 real roots, one of those roots must be a duplicate root, meaning there are really 3 real roots.

This is not true.

For example, consider:

[math]p(x) = x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)[/math]

What IS true, is if a real cubic polynomial has 2 real roots, the third root must be "self-conjugate", and thus also real.

As MarkFL's post(s) indicate the sum of the squares of each root is a symmetric polynomial in each root, and so must be expressible in terms of elementary symmetric polynomials of the three roots. This symmetric polynomial is of degree 2, and so is expressible in terms of elementary polynomials of degree 2 or less...and we KNOW what these evaluate to...the coefficients of our original polynomial!

That is, if:

[math]x^3 + x + 12 = (x - r_1)(x - r_2)(x - r_3)[/math]
[math]= x^3 - (r_1+r_2+r_3)x^2 + (r_1r_2+r_2r_3+r_1r_3)x - r_1r_2r_3[/math]

Then:

[math]r_1+r_2+r_3 = 0[/math]
[math]r_1r_2+r_2r_3+r_1r_3 = 1[/math]
[math]r_1r_2r_3 = -12[/math].

So since [math]r_1^2 + r_2^2 + r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2+r_2r_3+r_1r_3)[/math],

[math]r_1^2 + r_2^2 + r_3^2 = 0^2 - 2(1) = -2[/math].

Since real squares are always non-negative, at least one (and thus at least two) of our roots must be non-real.

Then it *is* true that since a real cubic (by the opening remarks of ILikeSerena's post) always has at least one root, we know we have exactly one real root, and it is easy to show (see above) that this root lies between -3 and -2.

Taking [math]r_1 = \alpha[/math], and from the fact that [math]r_3 = \overline{r_2}[/math], we see from:

[math]r_1r_2r_3 = -12 = \alpha|r_2|^2 = \alpha|r_3|^2[/math]

that [math]-12 = \alpha|r_2|^2 < -2|r_2|^2[/math], so that:

[math]|r_2|^2 < \frac{-12}{-2} = 6 \implies |r_2| < \sqrt{6}[/math].

Similarly, from [math]-3|r_2|^2 < \alpha|r_2|^2 = -12[/math], we obtain:

[math]|r_2|^2 > \frac{-12}{-3} = 4 \implies |r_2| > 2[/math], or:

[math]2 < |r_2| = |r_3| < \sqrt{6}[/math].

 

[I like Serena]

Re: Roots of polynomial equations

This is not true.

I think there is a miss communication of some sort.
As far as I can tell my statement is true and your counter example (which has 3 distinct roots and not 2) does not show why it wouldn't be.

 

[Deveno]

Re: Roots of polynomial equations

I think I see what you were saying: if a cubic has ONLY two real roots (and no others) one real root must be a duplicate.

The phrase "If it (a cubic polynomial) has two real roots..." is ambiguous, it might mean:

1) The cubic has at least two real roots (and possibly 3)
2) The cubic has exactly two real roots (and no others)
3) The cubic has exactly two real roots (and possibly a complex root)

(3), of course, cannot happen. I obviously took you to mean (1), when you meant (2).

As an aside, I also want to point out that even if one root (and thus, of course, two) is complex, it need not be (pure) imaginary, as evidenced by:

[math]p(x) = x^3 - 3x^2 + 4x - 2 = (x - 1)(x - (1+i))(x - (1-i))[/math]