In algebraic mathematics, polynomials are very important. Many equations have been used for the study of science. These equations will contain the variables and constants terms. Degree of the variable maybe 1, 2 or 3 and so on. Based on his degree equations van be classified as linear, quadratic, cubic, etc and so on. In this article, we will see the quadratic equations, its definition. Also, we will see the quadratic equation formula for getting the roots of the equation. Let us learn the interesting method.

**Quadratic Equation Formula**

**What is the Quadratic Equation?**

A quadratic equation is an equation having a second degree. It means it will contain at least one term in which the variable is squared. The standard form of this equation is:

ax²+ bx + c = 0

Where a, b, and c being constants or numerical coefficients

And x is an unknown variable.

One absolute rule is that the coefficient ‘a’ cannot be zero.

Due to the degree of 2, its variable X will have two possible values, which will satisfy the equation. We easily see that quadratic equations can represent many real-life situations. Now we will see the methods to solve it.

The name Quadratic comes from the word “quad” meaning square. Here we will develop the Quadratic Equation Formula to solve the quadratic equations. Let us start!

**Methods of Solving Quadratic Equations:**

Solving a quadratic equation means finding its two real roots which will be unique for a given equation. There are three popular methods for solving quadratic equations:

- Factorization
- Completing the square method
- Quadratic Equation Formula

The first and simplest method of solving quadratic equations is the factorization method. Certain quadratic equations can be factorized. These factors, if done correctly will give two linear equations in x. Hence, from these equations, we get the value of x. Let’s see an example and we will get to know more about it.

Here we will see the formula method.

**Quadratic Equation Formula:**

There are many equations that cannot be reduced using the factorization method. For such equations, a more powerful method is used. This method will work for every quadratic equation. This is the general quadratic equation formula.

It is defined as follows:

If ** ax² + bx + c = 0**

is a quadratic equation, then the value of x is given by the following formula.

Just put in the values of a, b and c, and then do the calculations.

The Quadratic equation Formula is having the values a, b, and c taken from the equation.

Here a, b, and c are just numbers. These are the numerical coefficients of the quadratic equation they’ve given you to solve.

X = \(\frac{[-b\pm (b^2-4 × a×c)^\frac{1}{2}]}{2×a}\)

Or

X = \(\frac{[-b\pm D^\frac{1}{2}]}{2×a}\)

Where,

D = \((b^2-4×a×c)^\frac{1}{2}\)

Where,

X | Variable |

a,b,c | Coefficients |

D | Discriminant |

Some cases of this equation are-

- If D= 0 hen both roots will be real and equal.
- If D<0 then no real root will exist for the given equation.
- Also, if D >0 then two real and different roots will exist.

Thus two roots will be

\(X_1=\frac{-b+D}{2×a}\)

\(X_2=\frac{-b-D}{2×a}\)

**Solved Examples**

Q. Solve the quadratic equation given as

- x²+3x-4=0

Finds its two roots.

**Solution:**

First ,** a = 1 , b = 3 , c= -4**

Then determine discriminant,

D = \((b^2-4×a×c)^\frac{1}{2}\)

D= \([3×3 – 4×1×(-4)]^\frac{1}{2}\)

D= 5

Now find first root.

\(x_1=\frac{-3+5}{2×1}\)

\(x_1 = 1\)

and then second root,

\(x_2=\frac{-3-5}{2×1}\)

\(x_2 = -4\)

Thus solution is x = –4 and x = 1.

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26

Hi

Same

yes