Quadratic equation is a second order polynomial equation. The general form of the quadratic equation in variable x, is $ax^{2}$ + bx + c = 0 where a $\neq$ 0. The solutions can be either both real, or both complex.

Quadratic equation is an equation of degree 2 and looks like
$ax^{2}$ + bx + c =0
a, b and c are real numbers called coefficients and 'a' cannot be zero.
x is the variable. This is the standard form of quadratic equation.
The quadratic formula made it appearance around 1100 AD and more or less we know from our high schools. The contributions of the ancient Indian Mathematicians to quadratic equations is highly significant.

Babylonian and Chinese contribution started a method, completing the square to solve basic problems involving area. In 17th century, Rene Descartes developed completing the square method into quadratic formula.
The solutions of a quadratic equation are called the roots of a quadratic equation.
Given below is the quadratic formula to solve quadratic equations in standard form

x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
The radical portion in the above formula $\sqrt{b^{2}-4ac}$ will determine the nature of the roots. The quantity in the radical sign $b^{2}$ - 4ac is known as the discriminant.

Given below is the table for three things regarding discriminant.

$b^{2}$ - 4ac > 0
Take the square root for the positive amount which leads to two different real roots to the equation.
$b^{2}$ - 4ac < 0 As we cannot take the square root of a negative number there will be no real roots.
$b^{2}$ - 4ac = 0 Square root of zero is zero so there will be equal real roots.

There are four most commonly used methods to solve quadratic equation. They are:
  1. Factoring
  2. Graphing
  3. Quadratic Formula
  4. Completing the Square
Given below are the steps to be considered while solving a quadratic equation by factoring.
  1. The given equation must be in standard form.
  2. Factor the left hand side as zero is on the right.
  3. Each factor should be set equal to zero.
  4. Solve to determine the roots.

Solved Example

Question: Solve  $2x^{2}$ + 7x + 3 = 0
Solution:
 
Step - 1 : The given equation is in the standard form.
                $2x^{2}$ + 7x + 3 = 0

Step - 2 : Consider $2x^{2}$ + 7x + 3 = 0
Factor the quadratic equation
$2x^{2}$ + 6x + x + 3 = 0
2x (x + 3) + 1 (x + 3) = 0
(x + 3) (2x + 1) = 0

Step - 3 : Set each factor to zero.
(x + 3) = 0  or (2x + 1) =0

Step - 4 : Determine the roots
(x + 3) = 0       $\Rightarrow$   x = - 3    or

(2x + 1) =0      $\Rightarrow$    x = $\frac{-1}{2}$

Thus x = -3  or x = $\frac{-1}{2}$
 

The standard form of the quadratic equation is $ax^{2}$ + bx + c = 0 and the graph of a quadratic equation will be a parabola.

A parabola can open up or down.
  • If the parabola opens up (a > 0), the lowest point is called the vertex.
  • If the parabola opens down (a < 0), the vertex is the highest point.

Given below are the steps for solving quadratic equation by graphing:
  1. For the given equation assign variables a, b and c.
  2. Check the parabola if it opens upwards or downwards.
  3. Find vertex:
  • x co-ordinate's minimum (maximum) point is given by
  • x = $\frac{-b}{2a}$
  • To get y co-ordinate plug in the value of x obtained into the quadratic equation.
4. Find the co-ordinates of the y-intercept by putting x = 0 in the given equation.
5. Find the co-ordinates of the x-intercept by putting y = 0 in the given equation and solve the quadratic equation.

Solved Example

Question: Solve  y = $x^{2}$ - 3x - 10 by graphing and find x.
Solution:
 
Step - 1 : Assigning the variables a = 1, b = -3, c = 10

Step - 2 : As a > 0, the graph of the given equation is a parabola with a minimum point and opens upwards.

Step - 3 : Find vertex:
x = $\frac{-b}{2a}$

x = $\frac{-(-3)}{2(1)}$

x = $\frac{3}{2}$        $\Rightarrow$ 1.5

So y = $(1.5)^{2}$ - 3(1.5) - 10
    y = -12.25
The Minimum point is (1.5, -12.25)

Step - 4 : To find y-intercept substitute x = 0 in the given equation

y = $(0)^{2}$ - 3(0) -10 = -10

(0, -10) is the y-intercept.

Step - 5 : Substitute y = 0 in the given equation and solve

$x^{2}$ - 3x -10

Factorize:  $x^{2}$ - 3x - 10

$x^{2}$ + 2x - 5x -10 = 0

x(x + 2) - 5(x + 2) = 0

(x + 2)(x - 5) = 0

=> x = 5, x = -2

The graph is given below :
Parabola

From the graph we can see that the values are  (-2, 0) and (5, 0) respectively .
 

Any quadratic equation of the form $ax^{2}$ + bx + c = 0, where a ≠ 0 can be solved for both real and
imaginary solutions using the quadratic formula:

x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Given below are the steps for solving quadratic equation by quadratic formula
  1. From the given equation find the coefficients of a, b and c.
  2. Plug in the values for a, b, and c into the quadratic formula.
  3. Simplify expression under the square root.
  4. Solve for x.

Solved Example

Question: Solve $x^{2}$ - 2x - 2 = 0 using quadratic formula.
Solution:
 
Step - 1 : The coefficients of a, b and c are a = 1, b = -2, c = -2

Step - 2 : The formula for quadratic equation is

x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Plug in the values of a, b and c into the quadratic formula.

x = $\frac{-(-2)\pm\sqrt{(-2)^{2}-4 \times 1 \times (-2)}}{2 \times 1}$

Step - 3 : Simplifying the expression:

x = $\frac{(2)\pm\sqrt{12}}{2}$

Step - 4 : Solve for x.

$x_{1}$ = $\frac{(2)+\sqrt{12}}{2}$            $\Rightarrow$   $x_{1}$ = 1 + $\sqrt{3}$

$x_{2}$ = $\frac{(2)-\sqrt{12}}{2}$              $\Rightarrow$   $x_{2}$ = 1 - $\sqrt{3}$
 

A polynomial equation in which one side is a perfect square trinomial is solved by taking square root on each side. This process of solving the equation is known as Completing the square.
Given below are the steps to be considered for solving the given expression in completing the square method:
  1. For the highest coefficient in the given term, divide each term by that value in the given expression.
  2. Move the constant to the right hand side.
  3. To find the value needed to make it a perfect square trinomial half the middle term (x), square it and add it to both sides of the equation.
  4. Factor the perfect square trinomial.
  5. Solve by taking square root on each side and make sure to consider both plus and minus results in the solution.

Solved Example

Question: Solve $x^{2}$ - 2x - 5 =0
Solution:
 
Simplify the equation by multiplying both sides with 5.

$x^{2}$ - 2x - 5 = 0
        
Move the constant to the right hand side.

$x^{2}$ - 2x = 5

Half the x- term coefficient, square it and add this value to both sides.

x-term coefficient = -2

Half the x-term coefficient = -1

After Squaring = 1

Add 1  to both sides we get  $x^{2}$ - 2x + 1 = 5 + 1

Simplifying the above
$x^{2}$ - 2x + 1 = 6

The perfect square for the left hand side is $(x-1)^{2}$ = 6

Taking square root on both sides   x - 1 = $\pm\sqrt{6}$

Therefore $x_{1}$ = 1 + $\sqrt{6}$
              $x_{2}$ = 1 - $\sqrt{6}$
 


Solved Example

Question: Solve  $x^{2}$ + 12x + 20 = 0
Solution:
 
Step 1: The given equation is in the standard form.
                $x^{2}$ + 12x + 20 = 0

Step 2: Consider $x^{2}$ + 12x + 20 = 0

Factor the quadratic equation

$x^{2}$ + 10x + 2x + 20 = 0

x(x + 10) + 2(x + 10) = 0

(x + 2) (x + 10) = 0

Step 3: Set each factor to zero.
(x + 2) = 0  or (x + 10) = 0

Step 4: Determine the roots
(x + 2) = 0       $\Rightarrow$   x = - 2   or
(x + 10) = 0      $\Rightarrow$   x =  - 10
Thus x = -2  or x = -10.