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## Introduction

A polynomial of the second degree is generally called a quadratic polynomial. In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing.

## What is a Quadratic Equation?

If p ( x ) is a quadratic polynomial, then p ( x ) = 0 is called a quadratic equation. Suppose f ( x ) is a quadratic polynomial, i.e. f ( x ) = ax2 + bx + c, a ≠ 0. Then f ( x ) = 0, i.e. ax2 + bx + c = 0 is called a quadratic equation.

Hence, we can say that the general form of a quadratic equation is given by

ax2 + bx + c = 0, where a, b, c, $\epsilon$ R and a ≠ 0

### Roots of a Quadratic Equation

Let p ( x ) = 0 be a quadratic equation. Then the zeros of the polynomial p ( x ) are called the roots of the equation p ( x ) = 0.

Thus x = α is a root of p ( x ) = 0 if and only if p ( α ) = 0.

It is important to note here that a quadratic polynomial may or may not have real zeros. In case a quadratic polynomial has real zeros, it can have at most two zeros. It follows from this that a quadratic equation can have at most two real roots. Finding the roots of a quadratic equation is known as solving the quadratic equation. Let us now understand different methods of solving a quadratic equation.

There are different methods using which we can find solutions to quadratic equations. The following method can be used to solve quadratic equations –

1. Solving Quadratic Equations Using Factorisation
2. Solving a Quadratic Equation by Completing the Square
3. Solving Quadratic Equations using the Discriminant
4. Graphical method for solving a Quadratic Equation

Let us understand these methods one by one. But, before that let us understand how to check whether a given equation is quadratic or not?

### Determining whether the given Equation is Quadratic or not

Let us consider two quadratic polynomials –

1. p (  x ) = x2 – 6 x  + 4
2. p (  x ) = x2 – 6 x3  + 4x + 5

In the first quadratic polynomial, we can see the highest order of the polynomial is 2, Hence,  p (  x ) = x2 – 6 x  + 4 is a quadratic polynomial.

In the second quadratic polynomial, we can see the highest order of the polynomial is 3, Hence,  p (  x ) = x2 – 6 x3  + 4x + 5 is not a quadratic polynomial.

Therefore, we can say that a polynomial is a quadratic equation if and only if the highest order of the equation is 2.

Now, we will discuss different methods to solve quadratic equations.

## Solving Quadratic Equations Using Factorisation

Let the quadratic equation be ax2 + bx + c = 0, where a, b, c, $\epsilon$ R and a ≠ 0. Let the quadratic polynomial, ax2 + bx + c be expressible as the product of two linear factors, say ( px + q ) and ( rx + s ) where p, q, r, s are real numbers such that p ≠ 0 and r ≠ 0.

Then,

ax2 + bx + c = 0

⇒ ( px + q ) ( rx + s ) = 0

⇒ px + q  = 0 or rx + s = 0

Solving these linear equations, we get the possible roots of the quadratic equation as

x = – $\frac{q}{p}$ and x = – $\frac{s}{r}$

Let us understand the above concept through an example.

Example

Solve the following quadratic equation by factorisation

x2 + 6x + 5 = 0

Solution

We have been given the quadratic equation, x2 + 6x + 5 = 0

Therefore,

x2 + 6x + 5 = 0

⇒ x2 + 5x + x + 5 = 0

⇒ x ( x + 5 ) +  ( x + 5 ) = 0

⇒ ( x + 5 ) ( x + 1 ) = 0

⇒ ( x + 5 ) = 0 or ( x + 1 ) = 0

⇒ x = – 5 or x = – 1

Thus, x  = – 5 and x = – 1 are the two roots of the equation x2 + 6x + 5 = 0

The above process is also known as solving the quadratic equation by splitting the middle term.

The following algorithm can be used to solve a quadratic equation using the above method.

1. Factorise the constant term of the given quadratic equation.
2. Express the coefficient of middle term as the sum or the difference of the factors obtained in the first step. Clearly, the product of these two factors will be equal to the product of the coefficient of x2 and the constant term.
3. Split the middle term in two parts obtained in the second step.
4. Factorise the quadratic equations by the factorisation method.

## Solving a Quadratic Equation by Completing the Square

We have learnt about the factorisation method to obtain the roots of a quadratic equation. Now, we shall learn the method of completing squares. The following algorithm may be used to obtain the roots of a quadratic equation by using the method of completing squares.

1. Obtain the quadratic equation. Let the equation be  ax2 + bx + c = 0.
2. Make the coefficients of x2 unity by dividing throughout by it, if it is not unity, obtain

x2 + $\frac{b}{a}$ x + $\frac{c}{a}$ = 0.

1. Shift the term ca on R.H.S to get x2 + $\frac{b}{a}$ x = – $\frac{c}{a}$
2. Add square of half of the coefficient of x i.e. ( $\frac{b}{2a}$ )2 on both sides to obtain

x2 +2 ( $\frac{b}{2a}$ ) x  + ( $\frac{b}{2a}$ )2 = ( $\frac{b}{2a}$ )2 – $\frac{c}{a}$

1. Write L.H.S as the perfect square of a binomial expression and simplify R.H.S to get

( x + $\frac{b}{2a}$ ) 2 = $\frac{b^2-4ac}{2a}$

1. Take square root on both sides, to get

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

1. Obtain the values of x by shifting the constant term $\frac{b}{2a}$   on R.H.S.

Let us understand the above algorithm using an example.

Example

Solve the quadratic equation 9x2 – 15x + 6 = 0 by using the method of completing squares.

Solution

We have been given the equation, 9x2 – 15x + 6 = 0. Let us use the above steps to solve this equation.

Therefore,

9x2 – 15x + 6 = 0

⇒ x2 –  $\frac{15}{9} x + \frac{6}{9}$ = 0

⇒ x2 –  $\frac{5}{3} x + \frac{2}{3}$ = 0

⇒ x2 –  $\frac{5}{3} x = – \frac{2}{3}$

⇒ x2 –  2 ( $\frac{5}{6} ) x + ( \frac{5}{6} )^2 = ( \frac{5}{6} )^2 – \frac{2}{3}$

⇒ $(x – \frac{5}{6} )^2 = \frac{25}{36} – \frac{2}{3}$

⇒ $(x – \frac{5}{6} )^2 = \frac{25-24}{36}$

⇒ $(x – \frac{5}{6} )^2 = \frac{1}{36}$

⇒ $x – \frac{5}{6} = ± \frac{1}{6}$

⇒ $x = \frac{5}{6} ± \frac{1}{6}$

⇒ $x = \frac{5}{6} + \frac{1}{6} or x = \frac{5}{6} – \frac{1}{6}$

⇒ x = 1 or x = $\frac{2}{3}$

Hence, x = 1 or x = $\frac{2}{3}$ are the two roots of the quadratic equation 9x2 – 15x + 6 = 0

## Solving Quadratic Equations using the Discriminant

If ax2 + bx + c = 0 is a quadratic equation, then the expression b2 – 4ac is known as the discriminant and is generally denoted by D.

The following theory of quadratic formula is used to define the method of solving a quadratic equation using the discriminant

The theory of quadratic equation formulae can be summarised as under to solve different types of problems on the quadratic equation. We know that the general form of a quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers (constants) and a ≠ 0, while b and c may be zero. Now, we have the summarised information about solving quadratic equations as below –

1. If ax2 + bx + c = 0 is a quadratic equation, then the expression b2 – 4ac is known as the discriminant and is generally denoted by D.
2. If α and β be the roots of the equation ax2 + bx + c = 0 (a ≠ 0) then
α + β =  – b/a = – [coefficient of x / coefficient of x2] and
αβ = c/a = [constant term / coefficient of x2]
3. The formula for the formation of the quadratic equation whose roots are given will be –
x2 – (sum of the roots)x + product of the roots = 0.
4. When a, b and c are real numbers, a ≠ 0 and discriminant is positive (i.e., b2 – 4ac > 0), then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real and unequal.
5. When a, b and c are real numbers, a ≠ 0 and discriminant is zero (i.e., b2 – 4ac = 0), then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real and equal.
6. When a, b and c are real numbers, a ≠ 0 and discriminant is negative (i.e., b2 – 4ac < 0), then the roots α and β of the quadratic equation ax2 + bx + c = 0 are unequal and imaginary. Here the roots α and β are a pair of the complex conjugates.
7. When a, b and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real, rational unequal.
8. When a, b and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax2 + bx + c = 0 are real, irrational and unequal.
9. When a, b and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax2 + bx + c = 0 are irrational.
10. In a quadratic equation with real coefficients has a complex root α + iβ then it has also the conjugate complex root α – iβ.
11. In a quadratic equation with rational coefficients has an irrational or surd root α + √β, where α and β are rational and β is not a perfect square, then it has also a conjugate root α – √β.

Let us understand the above concept using an example.

Example

Solve the quadratic equation 9x2  + 7x – 2 = 0

Solution

We have been given the equation, 9x2  + 7x – 2 = 0

Here, a = 9, b = 7 and c = -2

Now, we know that the discriminant D is given by

D = b2 – 4ac

Therefore,

D = 72 – 4 ( 9 ) ( – 2 ) = 49 + 72 = 121

Since 121 > 0 therefore, the given equation will have real roots.

Now, the roots of the equation are given by

$α = \frac{-b+\sqrt{D}}{2a} and β = \frac{-b- \sqrt{D}}{2a}$

Therefore,

α = $\frac{-7 + \sqrt{121}}{18} = \frac{-7+11}{18} = \frac{4}{18} = \frac{2}{9}$ and

β = $\frac{-7 – \sqrt{121}}{18} = \frac{-7-11}{18} = \frac{18}{18}$ =  – 1

Hence, $\frac{2}{9}$ and -1 are the two roots of the equation 9x2  + 7x – 2 = 0

## Graphical method for solving a Quadratic Equation

Plotting on a graph is another method of solving quadratic equations. The solution of the equation is obtained by reading the x-intercepts of the graph.

The following steps are used to solve a quadratic equation using graphs –

1. Given a quadratic equation, rewrite the equation by equating it to y or f(x)
2. Choose arbitrary values of x and y to plot the curve
3. Now graph the function.
4. Read the roots where the curve crosses or touches the x-axis.

Example

Solve the equation x2 + x – 3 = 0 by graphical method

Solution

We have been given the equation, x2 + x – 3 = 0

We will first find arbitrary values for the above equation. We get

We will now plot these values on a graph to get –

We can see that the x- intercepts are x = 1.3 and x = –2.3.

Therefore, the roots of the quadratic equation x2 + x – 3 = 0 are x = 1.3 and x = –2.3

It is important to note that there are three possibilities when solving quadratic equations by graphical method:

1. An equation has one root or solution if the x-intercept of the graph is 1.
2. An equation with two roots has 2 x -intercepts
3. If there is no x – intercepts, then an equation has no real solutions.

Let us understand how to use a graph for solving a quadratic equation.

## Key Facts and Summary

1. The general form of a quadratic equation is given by ax2 + bx + c = 0, where a, b, c, $\epsilon$ R and a ≠ 0.
2. A quadratic polynomial may or may not have real zeros. In case a quadratic polynomial has real zeros, it can have at most two zeros.
3. A polynomial is a quadratic equation if and only if the highest order of the equation is 2.
4. A quadratic equation can be solved by many methods that include factorization, completing the square, using the discriminant or using a graph.
5. If ax2 + bx + c = 0 is a quadratic equation, then the expression b2 – 4ac is known as the discriminant and is generally denoted by D.
6. If α and β be the roots of the equation ax2 + bx + c = 0 (a ≠ 0) then
α + β =  – b/a = – [coefficient of x / coefficient of x2] and αβ = c/a = [constant term / coefficient of x2]
7. The formula for the formation of the quadratic equation whose roots are given will be – x2 – (sum of the roots)x + product of the roots = 0.
8. Plotting on a graph is another method of solving quadratic equations. The solution of the equation is obtained by reading the x-intercepts of the graph. In a graph, if there is no x – intercepts, then an equation has no real solutions.