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Roots of polynomial equations 1

Erfan

New member
Jul 19, 2013
9
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

Administrator
Staff member
Feb 24, 2012
13,775
Re: Roots of polynomial equations

If we let the three roots be \(\displaystyle \alpha,\,z_1,\,z_2\) and write:

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

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

By Vieta, we have:

\(\displaystyle \alpha+z_1+z_2=-\frac{0}{1}=0\)

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

Hence:

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

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:

\(\displaystyle x^3+x=-12\)

For the moduli of the complex roots, consider:

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

Now, we may state (using Vieta again):

\(\displaystyle \alpha z_1z_2=-\frac{12}{1}=-12\)

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

Erfan

New member
Jul 19, 2013
9
Re: Roots of polynomial equations

If we let the three roots be \(\displaystyle \alpha,\,z_1,\,z_2\) and write:

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

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

By Vieta, we have:

\(\displaystyle \alpha+z_1+z_2=-\frac{0}{1}=0\)

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

Hence:

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

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:

\(\displaystyle x^3+x=-12\)

For the moduli of the complex roots, consider:

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

Now, we may state (using Vieta again):

\(\displaystyle \alpha z_1z_2=-\frac{12}{1}=-12\)

Putting this together with the polar form of the complex roots, what do you find?
I'm not really good at complex number .
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Roots of polynomial equations

What is:

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

Or, to avoid trigonometry, let:

\(\displaystyle z_1=a+bi\)

\(\displaystyle z_2=a-bi\)

Thus:

\(\displaystyle z_1z_2=(a+bi)(a-bi)\)

In both cases, recall:

\(\displaystyle i^2=-1\)

and for the polar form, recall:

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

zzephod

Well-known member
Feb 3, 2013
134
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.

.
 
Last edited:

zzephod

Well-known member
Feb 3, 2013
134
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 ,
\(\displaystyle f(x)=x^3 + x + 12\) is -ve at \(\displaystyle x=-3\) and +ve at \(\displaystyle x=-2\) hence \(\displaystyle f(x)\) has a root in the interval \(\displaystyle (-3,-2).\)

.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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 \(\displaystyle \alpha^2+z_1^2+z_2^2=0^2-2(1)=-2\) 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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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:

\(\displaystyle p(x) = x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)\)

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:

\(\displaystyle x^3 + x + 12 = (x - r_1)(x - r_2)(x - r_3)\)
\(\displaystyle = x^3 - (r_1+r_2+r_3)x^2 + (r_1r_2+r_2r_3+r_1r_3)x - r_1r_2r_3\)

Then:

\(\displaystyle r_1+r_2+r_3 = 0\)
\(\displaystyle r_1r_2+r_2r_3+r_1r_3 = 1\)
\(\displaystyle r_1r_2r_3 = -12\).

So since \(\displaystyle r_1^2 + r_2^2 + r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2+r_2r_3+r_1r_3)\),

\(\displaystyle r_1^2 + r_2^2 + r_3^2 = 0^2 - 2(1) = -2\).

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 \(\displaystyle r_1 = \alpha\), and from the fact that \(\displaystyle r_3 = \overline{r_2}\), we see from:

\(\displaystyle r_1r_2r_3 = -12 = \alpha|r_2|^2 = \alpha|r_3|^2\)

that \(\displaystyle -12 = \alpha|r_2|^2 < -2|r_2|^2\), so that:

\(\displaystyle |r_2|^2 < \frac{-12}{-2} = 6 \implies |r_2| < \sqrt{6}\).

Similarly, from \(\displaystyle -3|r_2|^2 < \alpha|r_2|^2 = -12\), we obtain:

\(\displaystyle |r_2|^2 > \frac{-12}{-3} = 4 \implies |r_2| > 2\), or:

\(\displaystyle 2 < |r_2| = |r_3| < \sqrt{6}\).
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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:

\(\displaystyle p(x) = x^3 - 3x^2 + 4x - 2 = (x - 1)(x - (1+i))(x - (1-i))\)