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**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 **ax ^{2} + 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:, 8x

As Example:

^{2}+ 5x – 10 = 0 is a quadratic equation.

**Root of quadratic equation:**

Root of a quadratic equation ax^{2} + bx + c = 0, is defined as real number α, if aα^{2} + bα + c = 0. The zeroes of the quadratic polynomial and the roots of the quadratic equation ax^{2} + 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 ax^{2} + bx + c, a ≠ 0, into a product of two linear factors,

then we can find the roots of the quadratic equation ax^{2} + 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.

**x ^{2 }+ 2mx + m^{2} = (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 **ax ^{2} + bx + c = 0**,

if the *determinant***b ^{2} – 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:

*Find the roots of the quadratic equation 6x*^{2}– x – 2 = 0.

**Solution:**

6x^{2} – x – 2

= 6x^{2} + 3x – 4x – 2

= 3x (2x + 1) – 2 (2x + 1)

= (3x – 2)(2x + 1)

The roots of 6x^{2} – 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:

*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

25x^{2} – 30x – 10 = 0

This can be also written as

**(5x) ^{2} – 2.(5x).3 + 3^{2} – 3^{2} – 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:

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

**Solution:**

In this equation **3x ^{2} – 5x + 2 = 0, **a = 3, b = -5, c = 2

let’s first check its

**determinant**which is

**b**which is 25 – 24 =

^{2}– 4ac,**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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