A polynomial is an algebraic expression consisting of many terms involving powers of the variable. The general form of the polynomial is

f(x) = a0xn + a1xn-1 + a2xn-2 + a3xn-3 + …+ an where ao, a1, a2, …. an are rational numbers and n is non-negative. The highest power of x is called the degree of the equation.

e.g. x6+ x3+1 is an equation with degree six.

If the degree is 1, then the polynomial is referred to as a linear equation. Similarly, if the degree is 2, then the polynomial is referred to as a quadratic equation.

The value of x for which the polynomial f(x) reduces to zero is called the root of the equation. Graphically, it is the point at which the graph of f(x) cuts the X-axis.

Properties of roots

1. The number of roots will depend on the degree of the equation. A polynomial of the nth degree will have n roots. For example, the equation x4 + 4x2+2 =0 has 4 roots. Note that the roots can be either real or imaginary. The total number of roots will be 4.

2. Imaginary roots occur in pairs. Thus, if 5+3i is one root of f(x), then 5-3i will also be a root of f(x).

3. Descartes Rule: This rule is used to find the maximum number of positive and negative real roots of a polynomial. Note that it does not provide the exact number of positive and negative real roots.

The number of positive roots of a polynomial is either equal to the number of sign changes between consecutive coefficients or it is less than it by an even number.

e.g. f(x) = x5 – x4 + 3x3 + 9x2 – x + 5. Notice that the sign changes from x5 to x4, from x4 to 3x3, from 9x2 to x and again from x to 5. Therefore, there are 4 sign changes. Hence, 4 is the maximum possible number of positive roots.

As per Descartes Rule, the number of positive roots could be 4 or 2 (=4-2) or 0 (=2-2). Basically, the number has to be counted down by 2 from the maximum possible number of roots.

Hence for f(x) = x5 – x4 + 3x3 + 9x2 – x + 5, there are either 4, 2 or 0 positive roots.

Similarly for negative roots, we need to check the number of sign changes for f(-x).

f(-x) = -x5 – x4 – 3x3 + 9x2 + x + 5

We can see that there is only 1 sign change (from 3x3 to 9x2). Hence the maximum possible number of negative real roots is 1. In this case, we don’t need to count down by 2 as doing so will result in a negative number. Hence, we know that f(x) contains only 1 negative root.

Summing up, f(x) has 4,2 or 0 positive roots and exactly 1 negative root.

  1. Corollaries of Descartes Rule:

a. If the coefficients of f(x) are all positive then the equation has no positive root. This is because there are no sign changes if all coefficients are positive.

b. If the coefficients of even powers of x are all of one sign and the coefficients of odd powers of x are all of opposite signs then the equation has no negative real root. Again, this follows from the sign change rule of Descartes.

c. If the equation contains only even powers of x and all the coefficients are of the same sign, then the equation will have no real roots.

d. If the equation contains only odd powers of x and all the coefficients are of the same sign, then the equation will only have 0 as a real root.

5. Sum and product of roots:

For a polynomial f(x) = a0xn + a1xn-1 + a2xn-2 + a3xn-3 + …+ an, let α1, α2, α3…. αn be the roots.

Then,

Sum of the roots (α12+ α3+ …+αn) = – a1 / a0

Sum of the products of the roots taken two at a time (α1α2+ α2α3+….) = a2/a0

Sum of the product of the roots taken three at a time (α1α2 α3+ α2α3 α4+….)  = – a3 / a0

…..

Product of the roots (α1 α2 α3…. αn ) = (-1)n an / a0

 

Let’s look at a few examples:

Q.1. If the equation x3 – ax2 + bx – a =0 has three real roots, then it must be the case that (CAT 2000)

a. b=1     b. b≠1       c. a=1         d. a≠1

Answer:  We can write the above equation as x2 (x – a) + bx – a =0

If b = 1 then we can take (x-a) as common.

Let’s take b=1 and check.

x2 (x – a) + x – a =0

(x-a) (x2+1)=0

Hence, x= a is one root. The other roots are x2 =-1 => the other two roots are imaginary.

But the question says that the equation has three real roots. Hence, the assumption that b= 1 is wrong. Hence, b≠1 which is option b.

Q.2. The number of roots common between the two equations x3 + 3x2 + 4x + 5 = 0 and x3 + 2x2 + 7x + 3 = 0 is (CAT 2003)

a. 0      b. 1      c. 2      d. 3

Answer: Equate both the equations to find the intersection points.

x3 + 3x2 + 4x + 5 = x3 + 2x2 + 7x + 3

=> x2-3x+2=0

=> x=2 or x=1

However, don’t make the mistake of taking them for the common roots. These are just the intersection points. We need to check if any of these is a root of both equations.

Substituting x=2 in the first equation will give us some positive value and hence it is not equal to 0. Hence, 2 cannot be a root of the first equation.

Similarly, substituting x=1 in the first equation will give us some positive value and hence it is not equal to 0. Hence, 1 cannot be a root of the first equation.

Hence, neither of the intersection points are roots. Hence, there is no root common to both the equations. Hence option a is the answer.

 

Here’s one for you to solve:

Q.3. If the roots of the equation x3 – ax2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b? (CAT 2008)

a. -1/√3       b. –1    c. 0     d. 1     e. 1/√3

 

 

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