How to Solve Quadratic Equations

Mathematical problems are represented in different type of equations and many equations are in the polynomial form of various order. Polynomial equation is formed with variable derivatives and constant integers which are co-relate to each other by arithmetic operators. In algebra equations plays the major role. Equations are the mathematical statements that define the equality of statements. In algebra the polynomial equations are of multiple forms as they are defined on the bases of degree. The equations with degree one is defined as linear equations, with degree two as quadratic and so on.

Friends today, we will discuss that how to solve Quadratic equations with ease? The standard form of Quadratic equation consist of only one variable derivatives whose highest degree is 2 that's why Quadratic equations are known as 2nd order equations.

Standard form is given as ax2 + bx + c = 0, here 'a', 'b' and 'c' are constants which are required to evaluate the solution of any quadratic equation. For solving quadratic a standard quadratic formula is used which gives two roots of the equation as its solution. The roots are given as:

First root = (-b + ( √(b2 - 4 ac) )) / 2 a.

Second root = (-b - √( b2 - 4 ac) )) / 2a.

Let us take an example of quadratic equation and see its evolution with the use of quadratic formula:

2x2 + 2x + 1 = 0 ( already in standard form of equation so no need to convert just recognize the coefficients) by comparing this equation with standard equation. On doing this we get,

a= 2, b = 2 and c = 1

now, put these value in quadratic formula:

First root = (- 2 + √( 22 - 4(2)(1))) / 2 (2) = ( - 2 + √( 4 - 8)) / 4 = -2 - √4 )/4 = -1

Second root = (- 2 - √( 22 - 4(2)(1))) / 2 (2) = (- 2 - √(4 - 8)) / 4 = -2 +√4 )/4 = 0

So the roots are (-1, 0)

To solve Algebra problems like complex quadratic students can use the quadratic equation solver, which is an online tool for fast solution of this type of equations.

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