Sum and product of roots with Examples and FAQs

Sum of roots and product of roots of quadratic polynomial

There are two values of the variable for which the quadratic polynomial has zero value, on the whole, these values are the roots of a quadratic polynomial. There exists a relation between the sum and product of roots with the coefficients of variables in the polynomial. This method makes it easy to calculate the sum and product of roots without actually knowing the roots.

Polynomial

Polynomials are expressions that are formed by variables.

A polynomial in one variable, “x” of degree “n” is in the form:

f(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x^{1}+a_{n} x^{0}

where a_{0}, a_{1}, a_{2}, \ldots, a_{n-1}, a_{n}are coefficients of the ‘n’ terms of the polynomial and
a_{0} \neq 0.

Quadratic Polynomial

A polynomial with the highest degree ‘2’ is known as a quadratic polynomial, i.e., the highest power of the variable is 2.

For example f(x)=4 x^{2}+2 x-1 is a quadratic polynomial as the highest degree of variable ‘x’ is 2.

Roots of a Quadratic Polynomial

The roots of a polynomial are the values of the variable for which the polynomial as a whole becomes zero-valued.

For example, consider the polynomial: f(x) = x – 20 The root of the polynomial is 20 because when x = 20, f(x) = 0.

Consider a quadratic polynomial:

f(x)=x^{2}-16, the roots of the polynomial are ±4.

When x=\pm 4, f(x)=16-16=0.

Note: When the coefficient of all the terms and the constant are zero then the polynomial is known as a zero polynomial.

A polynomial of degree n will have ‘n’ number of roots which are also known as zeroes. Thus, a quadratic polynomial has 2 roots, whereas a cubic polynomial has 3 roots, similarly, a biquadratic polynomial has 4 roots, and so on.

General form and roots formula of quadratic polynomial

The quadratic polynomial in the general form can be represented as:

f(x)=a x^{2}+b x+c, where a\neq 0.

The formula to determine the roots of a quadratic is:

x=\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}

For the polynomial:

f(x)=a x^{2}+b x+c, \text { where } a \neq 0

let\alpha and\betaare the roots of the quadratic polynomial. Then(x-\alpha) and(x-\beta) are the factors of the polynomial f(x). Therefore:

f(x)=a x^{2}+b x+c=p(x-\alpha)(x-\beta),where p is a constant

\therefore a x^{2}+b x+c=p\left(x^{2}-(\alpha+\beta) x+\alpha \beta\right)

a x^{2}+b x+c=p x^{2}-p(\alpha+\beta) x+p \alpha \beta

Comparing coefficients of x^{2}, xand constant terms on both sides of the equation we get:

\mathrm{a}=\mathrm{p}, \mathrm{b}=-p(\alpha+\beta) \text { and } \mathrm{c}=p \alpha \beta

\therefore(\alpha+\beta)=-\frac{b}{p}=-\frac{b}{a}=-\frac{(\text { coefficient of } x)}{\text { coefficient of } x^{2}}

And, \alpha \times \beta=\frac{c}{p}=\frac{c}{a}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}

Hence we can say that for the polynomial:

f(x)=a x^{2}+b x+c, \text { where } a \neq 0, \alpha and \betaare the roots of the quadratic polynomial then:

1. The sum of the roots, \boldsymbol{\alpha}+\boldsymbol{\beta}=-\frac{b}{a}=-\frac{(\text { coefficient of } x)}{\text { coefficient of } x^{2}}

2. The product of the roots, \boldsymbol{\alpha} \times \boldsymbol{\beta}=\frac{c}{a}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}

EXAMPLES 

Example 1: A quadratic polynomial f(x)=x^{2}-9 x+20 is given. Calculate the sum and product of the roots using the coefficients, and then verify your answer by finding the roots of the quadratic polynomial by factorization.

Solution: 

Given polynomial is f(x)=x^{2}-9 x+20.

Let the roots of the polynomial f(x) be \alpha and \beta

Then we have sum of roots, \alpha+\beta=-\frac{(\text { coefficient of } x)}{\text { coefficient of } x^{2}}=-\frac{(-9)}{1}=9.

And product of roots \alpha \times \beta=\frac{\text { constant term }}{\text { coefficient of } x^{2}}=\frac{20}{1}=20.

f(x)=x^{2}-9 x+20

         =x^{2}-5 x-4 x+20

         =x(x-5)-4(x-5)

         = (x - 5)(x - 4)

⇒ f(x)=0 at x=5, 4.

Hence the roots of f(x) are 5 and 4.

Now sum of roots = 5 + 4 = 9, and

product of roots = 5 × 4 = 20, hence verified.

Example 2: The sum and product of the roots of a quadratic polynomial f(x) are 5 and 6 respectively. Also, f(4) = 2. Determine the polynomial f(x).

Solution:

Let the polynomial be f(x)=a x^{2}+b x+c, where a\neq 0.

Let the roots of the polynomial f(x) be \alphaand\beta.

Given that f(4) = 2:

⇒f(4)=a(4^{2})-b(4)+c

\therefore f(4)=16a-4b+c ….(1)

It is also given that the sum of the roots is 5 and the product of roots is 6.

\therefore\left(-\frac{b}{a}\right)=5 \text { and } \frac{c}{a}=6

Hence, b = -5a and c = 6a.

Substituting values of b and c in (1)

   16a + 4(-5a) + 6a = 2

⇒ 16a – 20a + 6a = 2

⇒ 2a = 2

∴ a = 1

⇒ b = -5a = -5 and c = 6a = 6

Hence, the quadratic polynomial is f(x)=x^{2}-5 x+6.