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.

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:
1. 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:
1. 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.

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