Nature of Roots and Nature of Coefficients of Polynomial Equations

**Nature of Roots and Nature of Coefficients of Polynomial
Equations**

For a quadratic equation with real coefficients, if *α** *+ *i**β** *is a root, then *α** *− *i**β** *is also a root. In this section we shall
prove that this is true for higher degree polynomials as well.

We now prove one of the very important theorems in the theory of
equations.

If a complex number *z*_{0} is a root of a polynomial equation with
real coefficients, then its complex conjugate _{0} is also a root.

**Proof**

Let *P *( *x*) = *a*_{n}* x*^{n}* *+ *a*_{n}_{-1} *x*^{n}^{-1} +...+ *a*_{1} *x *+ *a*_{o}* *= be a polynomial
equation with real coefficients.

Let *z*_{0} be a root of this polynomial equation. So, *P*( *z*_{0}) = 0. Now

That is P(0 ) = 0 ; this implies that whenever z_{0} is a root (i.e. P( z_{0} )=0), its conjugate 0 is also a root.

If one asks whether 2 is a complex number, many students
hesitate to say “yes”. As every integer is a rational number, we know that
every real number is also a complex number. So to clearly specify a complex
number that is not a real number, that is to specify numbers of form *α *+
*iβ *with *β ±* 0 , we use the term **“non-real complex
number”****.
**Some authors call such
a number an **imaginary number**.

Remark 1

Let *z*_{0} = *α *+ *iβ *with *β *± 0 . Then 0 = *α *- *iβ *. If *α *+ *iβ *is a root of a polynomial
equation *P*(*x*) = 0 with real coefficients, then by Complex
Conjugate Root Theorem, *α *- *iβ *is also a root of *P*(*x*)
= 0 .

Usually the above statement will be stated as **complex roots occur in
pairs****;
**but actually it means
that **non-real complex roots
or imaginary roots occur as conjugate pairs****, ****being the coefficients
of the polynomial equation are real****.**

Remark 2

From this we see that any odd degree polynomial equation with
real coefficients has at least one real root; in fact, the number of real roots
of an odd degree polynomial equation with real coefficients is always an odd
number. Similarly the number of real roots of an even degree polynomial
equation with real coefficients is always an even number.

**Example 3.8**

Find the monic polynomial equation of minimum degree with real
coefficients having 2 - √3 *i* as a root.

Since 2 - √3*i* is a root of the required polynomial equation
with real coefficients, 2 + √3*i* is also a root. Hence the sum of the
roots is 4 and the product of the roots is 7 . Thus x^{2} - 4x + 7 = 0 is the
required monic polynomial equation.

If we further restrict the coefficients of the quadratic equation *ax*^{2} + *bx *+ *c *= 0 to be rational, we
get some interesting results. Let us consider a quadratic equation *ax*^{2} + *bx *+ *c *= 0 with *a *, *b*,
and *c *rational.
As usual let Δ = *b*^{2} − 4*ac *and let *r*_{1}* *and *r*_{2}* *be the roots. In this
case, when Δ = 0 , we have *r*_{1} = *r*_{2} ; this root is not
only real, it is in fact a rational number.

When Δ is positive, then no doubt that √Δ exists in **R** and we get two distinct real
roots. But √Δ will be a rational
number for certain values of a, b, and c , and it is an irrational number for
other values of a, b , and c .

If √Δ is rational, then both r_{1} and r_{2} are rational.

If √Δ is irrational, then both r_{1} and r_{2} are irrational.

Immediately we have a question. If Δ > 0 , when will √Δ be rational and when
will it be irrational? To answer this question, first we observe that Δ is rational, as the
coefficients are rational numbers. So Δ = *m*/*n* for some positive
integers *m* and *n* with (*m, n*) = 1 where (*m, n*)
denotes the greatest common divisor of *m* and *n*. It is now easy to
understand that √Δ is rational if and only if both *m *and *n* are perfect
squares. Also, √Δ is irrational if and only if at least one of m and n is not a
perfect square.

We are familiar with irrational numbers of the type p + √q where p and q are
rational numbers and √q is irrational. Such numbers are called **surds**. As in the
case of imaginary roots, we can prove that if p + √q is a root of a
polynomial, then p - √q is also a root of the same polynomial equation, when all
the coefficients are rational numbers. Though this is true for polynomial
equation of any degree and can be proved using the technique used in the proof
of imaginary roots, we state and prove this only for a quadratic equation in
Theorem 3.3.

Before proving the theorem, we recall that if *a* and *b*
are rational numbers and *c* is an irrational number such that *a + bc*
is a rational number, then b must be 0 ; further if a + *bc* = 0 , then a
and b must be 0.

For instance, if a + *b*√2 ∈ **Q**, then b must
be 0 , and if a + b√2 = 0 then a = b = 0 . Now we state and prove a general
result as given below.

Let *p *and *q *be rational numbers such
that √*q *is irrational. If *p *+ √*q *is a root of a quadratic equation
with rational coefficients, then *p *− √*q *is also a root of the same equation.

**Proof**

We prove the theorem by assuming that the quadratic equation is
a monic polynomial equation.

The result for non-monic polynomial equation can be proved in a
similar way.

Let p and q be rational numbers such that √q is irrational. Let p
+ √q be a root of the equation *x*^{2} + *bx* + c = 0 where b and c are rational
numbers.

Let α be the other root. Computing the sum of the roots, we get

α + p + √q = -*b*

and hence α + √*q* = -b - p ∈ **Q**. Taking -b - p
as s , we have α + √q = s .

This implies that

α = s - √q.

Computing the product of the roots, gives

(s -√q )( p + √q) ) = c

and hence (sp - q) + (s - p) = c ∈ **Q**. Thus s - p = 0
. This implies that s = p and hence we get α = p – √q. So, the other root is p
- √q.

The statement of Theorem 3.3 may seem to be a little bit
complicated. We should not be in a hurry to make the theorem short by writing “**for a polynomial
equation with rational coefficients, irrational roots occur in pairs***”*. This is not true.

For instance, the equation *x** ^{3}* - 2 = 0 has only one
irrational root, namely.

Find a polynomial equation of minimum degree with rational
coefficients, having 2^{ }- √3 as a root.

Since 2 - √3 is a root and the coefficients are rational numbers,
2 + is also a root. A required polynomial equation is given by

* *^{2} - (Sum of the roots) *x* + Product of the roots = 0

and hence

*x*^{2} - 4*x *+1 = 0

is a required equation.

We note that the term “rational coefficients” is very important; otherwise, x - (2 - √3) = 0 will be a polynomial equation which has 2 - √3 as a root but not 2 + √3. We state the following result without proof.

**Let p and q be rational numbers so that √p and √q are irrational
numbers; further let one of √p and √q be not a rational multiple of the
other. If √p + √q is a root of a polynomial equation with rational
coefficients, then √p - √q , - √p + √q , and - √p - √q are also roots of the
same polynomial equation.**

Form a polynomial equation with integer coefficients with as a root.

Since is a root, x – is a factor. To remove the outermost square root, we take x + as another factor and find their product

Still we didn’t achieve our goal. So we include another factor x^{2} + [√2/√3] and get the
product

So, 3*x*^{4} - 2 = 0 is a required polynomial equation with the
integer coefficients.

Now we identify the nature of roots of the given equation without
solving the equation. The idea comes from the negativity, equality to 0,
positivity of Δ = *b*^{2} - 4*ac*.

If all the coefficients of a quadratic equation are integers,
then Δ is an integer, and when it is positive, we have, √Δ is rational if, and
only if, Δ is a perfect square. In other words, the
equation *ax*^{2} + *bx *+ *c *= 0 with integer coefficients has rational roots, if, and only
if, Δ is a perfect square.

What we discussed so far on polynomial equations of rational
coefficients holds for polynomial equations with integer coefficients as well.
In fact, multiplying the polynomial equation with rational coefficients, by a
common multiple of the denominators of the coefficients, we get a polynomial
equation of integer coefficients having the same roots. Of course, we have to
handle this situation carefully. For instance, there is a monic polynomial
equation of degree 1 with rational coefficients having 1/2 as a root, whereas
there is no monic polynomial equation of any degree with integer coefficients
having 1/2 as a root.

Show that the equation 2*x*^{2} - 6*x *+ 7 = 0
cannot be satisfied by any real values of *x.*

∆= *b*^{2} − 4*ac *= −20 < 0 . The roots are imaginary numbers.

**Example 3.12**

If *x*^{2} + 2 (*k *+ 2)*x *+ 9*k *= 0 has equal roots,
find *k*.

**Solution**

Here Δ = *b*^{2} - 4*ac *= 0 for equal roots.
This implies 4 (*k *+ 2)^{2} = 4 (9) *k *.This implies *k *=
4 or 1.

**Example 3.13**

Show that, if *p*, *q*, *r* are rational, the
roots of the equation *x*^{2} - 2 *px *+ *p*^{2} - *q*^{2} + 2*qr *- *r*^{2} = 0 are rational.

**Solution**

The roots are rational if Δ = *b*^{2} - 4*ac *=
(-2 *p*)^{2} - 4 ( *p*^{2} - *q*^{2} + 2*qr *- *r*^{2} ) .

But this expression reduces to 4 (*q*^{2} - 2*qr *+ *r*^{2} ) or 4 (*q *- *r
*)^{2} which is a
perfect square. Hence the roots are rational.

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12th Mathematics : UNIT 3 : Theory of Equations : Nature of Roots and Nature of Coefficients of Polynomial Equations |

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