Polynomial Equations with no Additional Information

**Reciprocal Equations**

Let α be a solution of the equation.

2x^{6} - 3x^{5} + √2x^{4} + 7x^{3} + √2x^{2} - 3x + 2 = 0. ... (1)

Then α ¹ 0 (why?) and

2α^{6} - 3α^{5} + √2α^{4} + 7α^{3} + √2α^{2} - 3α + 2 = 0.

Substituting 1/α for x in the left side of (1), we get

Thus 1/α is also a solution of (1). Similarly we can see that if α
is a solution of the equation

*x*^{5} + 3*x*^{4} - 4*x*^{3} + 4*x*^{2} - 3*x* – 2 = 0 ... (2)

then 1/α is also a solution of (2).

Equations (1) and (2) have a common property that, if we replace
*x *by 1/x in the equation and write it as a polynomial equation, then we
get back the same equation. The immediate question that flares up in our mind
is “Can we identify whether a given equation has this property or not just by
seeing it?” Theorem 3.6 below answers this question.

**Definition 3.1**

A polynomial *P*(*x*) of degree *n *is said to be
a reciprocal polynomial if one of the following conditions is true:

A polynomial *P*(*x*) of degree *n *is said to be
a reciprocal polynomial of Type I if *P*(*x*) = called a
reciprocal equation of Type I.

A polynomial *P*(*x*) of degree *n *is said to be
a reciprocal polynomial of Type II if *P*(*x*) = - called a
reciprocal equation of Type II.

A polynomial equation *a*_{n}* x*^{n}* *+ *a*_{n}_{−}_{1} *x*^{n}^{−}^{1} + *a*_{n}_{−}_{2} *x*^{n}^{−}^{2} +….+ *a*_{2} *x*^{2} + *a*_{1} *x *+ *a*_{0} = 0 , (*a*_{n}*≠ *0) is a reciprocal
equation if, and only if, one of the following two statements is true:

(i) *a*_{n}* *= *a*_{0} ,* a*_{n}_{-1} = *a*_{1} , *a*_{n}_{-2} = *a*_{2 }….

(ii) *a*_{n}* *= - *a*_{0} , *a*_{n}_{-1} = - *a*_{1} , *a*_{n}_{-2} = - *a*_{2} , …

Consider the polynomial equation

*P*(*x*) = *a*_{n}* x*^{n}* *+ *a*_{n}_{-1} *x*^{n}^{-1} + *a*_{n}_{-2} *x*^{n}^{-2} +...+ *a*_{2} *x*^{2} + *a*_{1} *x *+ *a*_{0} = 0 . … (1)

Replacing x by 1/x in (1), we get

Multiplying both sides of (2) by *x** ^{n}*, we get

Now, (1) is a reciprocal equation ⇔ P(*x*) = ± x^{n} P (1/*x*) ⇔ (1) and (3) are same .

Let the proportion be
equal to λ. Then, we get a_{n}/a_{0} = λ and a_{0}/a_{n} = λ . Multiplying
these equations, we get *λ*^{2} = 1. So, we get two cases *λ *= 1 and *λ
*= -1 .

**Case (i) :**

*λ *= 1 In this case, we have *a*_{n}* *= *a*_{0} , *a*_{n}_{−1} = *a*_{1} , *a*_{n}_{−2} = *a*_{2} , …..

That is, the coefficients of (1) from the beginning are equal to
the coefficients from the end.

**Case (ii) :**

*λ *= −1 In this case, we have *a*_{n}* *= −*a*_{0} , *a*_{n}_{−1} = −*a*_{1} , *a*_{n}_{−2} = −*a*_{2} , …..

That is, the coefficients of (1) from the beginning are equal in
magnitude to the coefficients from the end, but opposite in sign.

Reciprocal equations of Type I correspond to those in which the
coefficients from the beginning are equal to the coefficients from the end.

For instance, the equation 6*x*^{5} + *x*^{4} − 43*x*^{3} − 43*x*^{2} + *x *+ 6 = 0 is of type I.

Reciprocal equations of Type II correspond to those in which the
coefficients from the beginning are equal in magnitude to the coefficients from
the end, but opposite in sign.

For instance, the equation 6*x*^{5} − 41*x*^{4} + 97*x*^{3} − 97*x*2 + 41*x *− 6 = 0 is of Type II.

(i) A reciprocal equation cannot have 0 as a solution.

(ii) The coefficients and the solutions are not restricted to be
real.

(iii) The statement “**If ***P*(*x*) = 0 **is a polynomial equation such that whenever ***α** ***is a root,** 1/*α* is also a root, then
the polynomial equation *P *( *x*) = 0 must be a reciprocal equation” is not true. For instance 2*x*^{3} − 9*x*^{2} + 12*x *− 4 = 0 is a polynomial
equation whose roots are 2, 2,1/2.

Note that *x*^{3} P( 1/*x*) ≠ ± P(x) and hence it is not a
reciprocal equation. Reciprocal equations are classified as Type I and Type II
according to a_{n-r} = a_{r} or a_{n-r} = -a_{r} , r = 0, 1, 2,...n. We state some results without proof :

·
For an odd degree reciprocal equation of Type I, *x *= −1 must be a solution.

·
For an odd degree reciprocal equation of Type II, *x *= 1 must be a solution.

·
For an even degree reciprocal equation of Type II, the middle
term must be 0 Further *x *= 1 and *x *= −1 are solutions.

• For an even degree reciprocal equation, by taking x + (1/*x*)
or x – (1/x) as y , we can obtain a polynomial equation of degree one half of
the degree of the given equation ; solving this polynomial equation, we can get
the roots of the given polynomial equation.

As an illustration, let us consider the polynomial equation

6*x*^{6} - 35*x*^{5} + 56*x*^{4} - 56*x*^{2} + 35*x* - 6 = 0

which is an even degree reciprocal equation of Type II. So 1 and
-1 are two solutions of the equation and hence x^{2} -1 is a factor of the
polynomial. Dividing the polynomial by the factor x^{2} -1, we get 6*x*^{4} - 35*x*^{3} + 62*x*^{2} - 35*x* + 6 as a
factor. Dividing this factor by x2 and rearranging the terms we get .
Setting u = ( *x* + 1/*x*) it becomes a quadratic polynomial as
6 (u^{2} - 2) - 35u + 62 which
reduces to 6u^{2} - 35u + 50 . Solving we obtain u = 10/3 , 5/2 . Taking *u* =
10/3 gives *x* = 3, 1/3 and taking *u *= 5/2 gives *x* = 2, 1/2.
So the required solutions are +1, -1, 2, 1/2 , 3, 1/3 .

Solve the equation 7*x*^{3} − 43*x*^{2} = 43*x *− 7.

The given equation can be written as 7*x*^{3} - 43*x*^{2} - 43*x *+ 7 = 0.

This is an odd degree reciprocal equation of Type I. Thus -1 is
a solution and hence *x *+1 is a factor.

Dividing the polynomial 7*x*^{3} - 43*x*^{2} - 43*x *+ 7 by
the factor *x *+1,we get 7*x*^{2} - 50*x *+ 7 as a quotient.

Solving this we get 7 and 1/7 as roots. Thus -1, 1/7 , 7 are the
solutions of the given equation.

**Example 3.28**

Solve the following equation: *x*^{4} −10*x*^{3} + 26*x*^{2} −10*x *+1 = 0.

**Solution**

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as

Let y = x + [1/*x*] . Then, we get

(y^{2} - 2) -10 y + 26 = 0 ⇒ y^{2} -10 y + 24 =
0 ⇒ ( y -6)( y - 4) =
0 ⇒
y = 6 or y = 4

**Case (i)**

y = 6 ⇒ x + (1/*x*)
= 6 ⇒ x = 3 + 2√2, x = 3 -√2 .

**Case (ii)**

y = 4 ⇒ x + (1/*x*)
= 4 ⇒ x = 2 + √3, x = 2 -√3

Hence, the roots are 3 ± 2√2, 2 ±√3

Tags : Definition, Theorem, Formulas, Solved Example Problems , 12th Mathematics : UNIT 3 : Theory of Equations

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12th Mathematics : UNIT 3 : Theory of Equations : Reciprocal Equations | Definition, Theorem, Formulas, Solved Example Problems

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