The algebraic equation of degree two is called the quadratic equation. There are various methods for solving quadratic equations. The term quadratic is a Latin word quadratus that means square. The quadratic equation is univariable. The quadratic formula, factoring, completing the square will help to solve the quadratic equation for real and complex roots.

## Standard Form of Solving Quadratic Equations

There are certain steps for solving quadratic equations. First, we have to write the given expression in the standard form.
The general form of the Quadratic Equation is ax² + bx + c = 0
Where x represents unknown variable
a, b are the coefficients of x² and x.
c is the constant.
The values of x that satisfy the equation are called the solution of the equation. The roots of the expression lie on the left-hand side of the equation. The solutions of the equations may or may not be distinct they may not be real.

### Methods of Solving Quadratic Equation

By factoring:
It may be possible to express the quadratic equation as a product (px + q)(rx + s) = 0. Solving quadratic equation by factoring by simple inspection to determine values p, q, r, s. If the given equation is quadratic then write it in zero factor property if px + q = 0 or rx + s = 0. Solving two linear equations provides the roots of the quadratic.

Students can solve the quadratic equations by using the quadratic formula. For solving quadratic equations we have to write the general form of the quadratic equation. Identify the values by comparing them with the standard form of the equation. Place the values in the quadratic formula and find the roots of the equation.
The formula for solving the quadratic equation is x = [-b ± √(b² – 4ac)]/2a

By completing the square:

Write the general form of the quadratic equation and then divide both sides by coefficient of x². Shift the constant term to the RHS of the equation and add the square of half of the coefficient pf x on both sides. Write the LHS as a complete square and simplify the equation.

Example 1.
Solve the equation x² = 81
Solution:
Given the equation x² = 81
x² – 81 = 0
(x + 9)(x – 9) = 0
x + 9 = 0
x = -9
x – 9 = 0
x = 9
The solutions are {-9, 9}

Example 2.
Solve the equation x² – x – 6 = 0 by using quadratic formula.
Solution:
Given the equation,
x² – x – 6 = 0
The formula for solving the quadratic equation is x = [-b ± √(b² – 4ac)]/2a
a = 1, b = -1, c = -6
x = [-b ± √(b² – 4ac)]/2a
x = [1 ± √((-1)² – 4(1)(-6))]/2(1)
x = [1 ± √1 + 24)]/2
x = [1 ± √25)]/2
x = [1 + 5]/2 = 6/2 = 3
x = 3
x = [1 – 5]/2 = -4/2 = -2
x = -2
Thus the solution set is {3, -2}

Example 3.
Solve the equation 2x² + 5x – 3 = 0 by using quadratic formula.
Solution:
Given the equation,
2x² + 5x – 3 = 0
The formula for solving the quadratic equation is x = [-b ± √(b² – 4ac)]/2a
a = 2, b = 5, c = -3
x = [-b ± √(b² – 4ac)]/2a
x = [-5 ± √((-5)² – 4(2)(-3))]/2(2)
x = [-5 ± √25 + 24)]/4
x = [-5 ± √49)]/4
x = [-5 + 7]/4 = 2/4 = 1/2 = 0.5
x = 0.5
x = [-5 – 7]/4 = -12/4 = -3
x = -3
Thus the solution set is {0.5, -3}

Example 4.
Solve the equation a² – 3a = 0
Solution:
Given the equation a² – 3a = 0
a(a – 3) = 0
a = 0
a – 3 = 0
a = 3
The roots of the solution is {0, 3}

Example 5.
Solve the equation x² + 2x – 3 = 0 by factoring.
Solution:
Given the equation x² + 2x – 3 = 0
The general form of the Quadratic Equation is ax² + bx + c = 0
a = 1, b = 2, c = -3
x² + 3x – 1x – 3 = 0
x(x + 3)-1(x + 3) = 0
(x + 3)(x – 1) = 0
x + 3 = 0
x = -3
x – 1 = 0
x = 1
Thus the roots of the solution is {1, -3}

### FAQs on Solving Quadratic Equations

1. What are the 4 methods of solving quadratic equations?

The 4 methods of solving a quadratic equation are factoring, using the square roots, completing the square, and the quadratic formula.

2. What is the best method for solving quadratic equations?