Quadratic Equation

Before studying about this topic let’s know the word “quadratic” came from “quadratus” means square.

An algebraic equation or polynomial equation with degree 2 is said to be a quadratic equation.

It is represented in terms of variable “x” as ax² + bx + c = 0. This form of representation is called standard form of quadratic equation.  where a, b, c are real numbers and the important thing is a must be not equal to zero.
As Example:, 8x² + 5x – 10 = 0 is a quadratic equation.

Root of quadratic equation:

Root of a quadratic equation ax² + bx + c = 0, is defined as real number α, if aα² + bα + c = 0. The zeroes of the quadratic polynomial and the roots of the quadratic equation ax² + bx + c = 0 are the same.

Solution of a Quadratic Equation by different methods:

1. By Factorisation:

First thing to keep in mind that If we can factorise ax² + bx + c, a ≠ 0, into a product of two linear factors,
then we can find the roots of the quadratic equation ax² + bx + c = 0 by equating each linear factor to zero.

2. By Completing Square:

The term completing the square in algebra is to form the given term in squared units by the use of algebraic identities.

x² + 2mx + m² = (x + m)²

By this algorithm, we can find the roots easily. One of the fact to remember that when square root is opened in number it uses simultaneously both + as well as – sign. Thus two roots is defined.

3. By use of quadratic formula:

In the standard quadratic equation ax² + bx + c = 0,

if the determinant b² – 4ac ≥ 0

then root of quadratic equation is given by quadratic formula as

x =\(\frac{-b±\sqrt{b^{2}-4ac}}{2a}\)

Examples related to quadratic equation

• Solution by factorization examples:

1. Find the roots of the quadratic equation 6x² – x – 2 = 0.

Solution:

6x² – x – 2
= 6x² + 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x² – x – 2 = 0 are the values of x so that (3x – 2)(2x + 1) = 0
so, 3x – 2 = 0 or 2x + 1 = 0,
i.e. x = \(\frac{2}{3}\) or x = \(\frac{-1}{2}\)

So, roots of equation are \(\frac{2}{3}\) , \(\frac{-1}{2}\).

• Solution by completing square examples:

2. Find the roots of the equation 5x 2 – 6x – 2 = 0 by the method of completing the square.

Solution:

To solve it we first multiply the equation throughout by 5
25x² – 30x – 10 = 0
This can be also written as
(5x)² – 2.(5x).3 + 3² – 3² – 10 = 0
(5x – 3)² – 9 – 10 = 0
(5x – 3)² = 19
5x – 3 = ±\(\sqrt{19}\)
5x = 3 ± \(\sqrt{19}\)
x = \(\frac{3 ± \sqrt{19}}{5}\)

So, the roots of equation are \(\frac{3 + \sqrt{19}}{5}\) and x = \(\frac{3 - \sqrt{19}}{5}\)

• Solution by Quadratic formula examples:

3. Find the roots of the quadratic equation, 3x² – 5x + 2 = 0 if it exists, using the quadratic formula.

Solution:

In this equation 3x² – 5x + 2 = 0, a = 3, b = -5, c = 2
let’s first check its determinant which is b² – 4ac, which is 25 – 24 = 1 > 0, thus the solution exists.
by applying quadratic formula x =\(\frac{-b±\sqrt{b^{2}-4ac}}{2a}\)
we have, x = \(\frac{5 ± \sqrt{1}}{6}\) = \(\frac{5 ± 1}{6}\)
i.e, x = 1 or x = \(\frac{2}{3}\)
so, the roots are \(\frac{2}{3}\), 1 etc.