Quadratic Polynomial

A quadratic polynomial is a polynomial in the form of ax²+bx+c, where a, b and c are said to be real numbers and a is not equal to zero.. When we try to equate a quadratic polynomial with a constant, the resultant equation is in the form of quadratic equation. Any equation in the form a(x) = b, where a(x) is a polynomial of degree 2 and b is a constant, is said to be a quadratic equation.

The standard form of a Quadratic Polynomial is ax²+bx+c=0. Here, all a, b and c are real numbers and a is not equal to zero. a is the coefficient of to the power 2. It is known as a quadratic coefficient. ‘b’ is the coefficient of x and is known as the linear coefficient. ‘c’ is the constant term.

Roots of a Quadratic Polynomial Equation

The roots of the quadratic equation for which the values of x are given as a quadratic equation are known as quadratic polynomial. If p is a root of the quadratic equation ax2+bx+c=0, then ap²+bp+c=0. A quadratic equation can have two different real roots, two equal roots and also real roots may not exist at all for them. In graphical representation, the roots of a quadratic equation can be said to be the points where the graph of the quadratic polynomial dissects the x-axis.

Quadratic Equation by Factorisation Method

Consider a quadratic equation 2×2−5x+6=0

⇒2x²−2x−3x+3=0

We further simplify the middle part of the equation by finding two numbers (-2 and -4) ina way that their addition is equal to the coefficient of x and their multiplication is equal to the multiplication of the coefficient of x2 and the constant.

(-2) + (-3) = (-5)

And (-2) × (-3) =6

2×2−2x−3x+6=0

2x(x−1)−3(x−1)=0

(x−1)(2x−3)=0

In this step, we have denoted the quadratic polynomial as the multiplication of its two factors.
Thus, x = 1 and x =3/2 are said to be the roots of the given quadratic plynomial equation.
This method of simplifying a quadratic equation is known as the factorisation method.

Quadratic Equation Using Quadratic Formula

The quadratic formula is applied to directly determine the roots of a quadratic equation from the standard structure of the equation.

For the quadratic equation ay²+by+c=0,

y= [-b± √(b2-4ac)]/2a

By replacing the values of a, b and c, we can easily find the roots of the given equation.

Solved example: If y² – 5y + 6 = 0 is the quadratic equation, find the roots.

Solution: Given, y² – 5y + 6 = 0 is the quadratic equation.

On comparing with standard quadratic equation we have;

ay² + by + c = 0

a = 1, b = -5 and c = 6

Since,

b2 – 4ac = (-5)2 – 4 × 1 × 6 = 25 – 24 = 1 > 0

Hence, the roots are real.

Using quadratic formula,

y = [-b ± √(b2 – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. y = (5 + 1)/2 and x = (5 – 1)/2

y = 6/2, x = 4/2

y = 3, 2

Therefore, the roots of quadratic equation are 3 and 2.

Quadratic Equation by Discriminant Method

In a quadratic polynomial equation in the form ax²+bx+c=0, the expression b²−4ac is said to be the discriminant which is denoted by the letter D, of that quadratic equation. The discriminant determines the nature of roots that will be derived from the quadratic polynomial, based on the coefficients of the said quadratic equation.