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Roots of Quadratic Equation

Today we are going to learn how to find roots of a quadratic equation. This is a very important topic and it is going to help you out in advanced topics as well. Most of the questions from other topics involve finding the roots of the quadratic equations. So, even if you know the method to solve the question but you are stuck at finding the roots, you are going to be so frustrated at yourself and you certainly do not want that. Let us learn it so we can make our lives easy.

General Form of a Quadratic Equation

The general form of a quadratic equation is given as:

ax2 + bx + c = 0

where a, b and c are real numbers and b and c can be zero but a cannot.

For example,

x2 + 4x + 4 = 0 is a quadratic equation with a = 1, b =4 and c = 4.

x2 = -9 is a quadratic equation with a = 1, b = 0 and c = 9.

x + 9 = 0, it is a linear equation and not a quadratic equation. The quadratic equation needs to have the maximum degree of 2.

Also know: What is a linear equation

Roots of the Quadratic Equation

Which values should we put in the quadratic equation so it gets satisfied i.e. ax2 + bx + c gets equal to zero. For example, if we put x = -2 in the equation x2 + 4x + 4 = 0, it gets satisfied. i.e.

x2 + 4x + 4 = 0

(-2)2 + 4(-2) + 4 = 0

4 – 8 + 4 = 0

0 = 0

Hence, -2 is the root of the equation.

The root of the polynomial is equal to the highest degree present in that polynomial. Therefore, there will be 2 roots for the quadratic equation as the highest degree or the power is 2. The root is the value at which the graph touches the x-axis. It is equivalent to finding the solution of a quadratic equation.

How do we find the roots of the quadratic equation? There are several ways but first, we are going to study the generic formula which can be used to find the solution of any quadratic equation if one exists and it is known as the quadratic formula.

The Quadratic formula

The formula (known as Shri Dharacharya formula) to find the roots of a quadratic equation is:

x = (-b ± √{(b2 – 4ac)) / 2a}

The ± sign shows that there can be two roots. Thus, if α and β are the two roots of the quadratic equation. Then,

α = -b + √{(b2 – 4ac)) / 2a}

and β = (-b – √{(b2 – 4ac)) / 2a}

Will there always be two roots? Can it be the case where there will be no roots or only one root? It sure can be. But how can we know that? We can know this by finding the nature of roots.

Nature of Roots of Quadratic Equation

The nature of the roots of a quadratic equation can be found by the discriminant. i.e.

discriminant, D =  b2 – 4ac

where a, b, and c are real numbers and a cannot be zero.

There can be three cases of the discriminant and three nature of roots.

Case I: b2 – 4ac > 0, D > 0

If the discriminant is positive i.e. greater than zero, then the quadratic equation will have two real and distinct or unequal roots.

Case II: b2 – 4ac < 0, D < 0

If the discriminant is negative i.e. less than zero, then the quadratic equation will have two imaginary and unequal roots. This means that the curve never touches the x-axis. Any quadratic equation that is above or below the x-axis will never cross the x-axis and thus will have imaginary roots.

Case III: b2 – 4ac = 0, D = 0

If the discriminant is equal to zero, then the quadratic equation will have two real and equal roots. This means the curve touches the x-axis at only one point. i.e.

α = β = -b2 / 2a

Also know: how to find the roots of a cubic equation 

Summary: How to find the roots of a quadratic equation using Shri Dharacharya formula 

Roots of a quadratic equation

Sum of Roots of a Quadratic Equation

If α and β are the two roots of the quadratic equation ax2 + bx + c = 0, then,

The sum of the roots =  α + β = -b/a

Product of Roots of a quadratic equation

If α and β are the two roots of the quadratic equation ax2 + bx + c = 0, then,

The product of the roots =  α×β = c/a

Let us solve some sample questions now.

Example 1: Find the nature of the roots of following quadratic equations:

a) -2x2 – 5x + 4 = 0

b) 25x2 + 50x + 25 = 0

c) x2 + 2x + 3 =0

Solution

The nature of roots can be found by the discriminant i.e.

discriminant = D = b2 – 4ac

a) -2x2 – 5x + 4 = 0

Here a = -2, b = -5 and c = 4.

D = (-5)2 – 4(-2)(4)

D = 25 + 32

D = 57

Since D > 0, then the roots of the equation are real and distinct. Moreover, as D is not a perfect square, the roots are irrational. If D > 0 and a perfect square, then the roots will be rational.

b) 25x2 +50x + 25 = 0

Here a = 25, b = 50 and c = 25.

D = (50)2 – 4(25)(25)

D = 2500 – 2500

D = 0

Since D = 0, then the roots of the equation are real and equal.

c) x2 + 2x + 3 = 0

Here a = 1, b = 2 and c = 3.

D = (2)2 – 4(1)(3)

D = 4 – 12

D = -8

Since D < 0, then the roots of the equation are imaginary and unequal. The curve (parabola) is above or below the x-axis.

Example 2: What are the roots of a quadratic equation 4x2 + 5x – 6?

Solution

The roots of the quadratic equation can be found by using the quadratic formula.

x = (-b ± √(b2 – 4ac)) / 2a

Here a = 4, b = 5 and c = -6

x = (-5 ± √((5)2 – 4(4)(-6))) / 2(4)

x = (-5 ± √(25 + 96)) / 8

x = (-5 ± √(121)) / 8

x = (-5 ± 11) / 8

Now,

α = (-5 + 11) / 8 and β = (-5 – 11) / 8

α = 6/8    β = -16/8

α = 3/4 and β = -2

The two roots are 3/4 and -2.

There are two other methods to solve the quadratic equations as well.

Factoring Quadratic Equations

The solution to the quadratic equation is obtained by factoring the equation in such a way so that it is converted into the form (x-a)(x-b) = 0. Then the roots of the equation will be a and b.

By expansion,

(x-a)(x-b) = 0

x2 -bx -ax + b2 = 0

x2 – (a+b)x + b2 = 0

which is a quadratic equation.

Hence, if the given quadratic equation can be written in the form x2 – (a+b)x + b2, we can convert it into (x-a)(x-b) = 0 and obtain the roots.

Example

Let us solve the above same problem and let us see if we can get the same answer.

4x2 + 5x – 6 = 0

Steps to to find the roots of a quadratic equation using factorization method

  • Take the product of the coefficient of x2 and the constant term. We get 4 × -6 = -24.
  • Find out the prime factors of 24. i.e. 24 = 2 × 2 × 2 × 3
  • Obtain a pair of values from these factors that on addition or subtraction give 5x. In this case, 8x – 3x = 5x. 8 is obtained by 2 × 2 × 2

Now, we can write the equation as

4x2 + 8x – 3x – 6 = 0

4x(x + 2) -3(x+2) =0

(x + 2)(4x-3)=0

Now,

x+2 = 0 and 4x – 3 = 0

x  = -2 and x = 3/4

which are the same values that we found above.

Can we solve every quadratic equation by factorization method?

Let us solve another problem.

5x2 + 5x – 6 = 0

Taking the product of the coefficient of x2 and the constant term. We get 5 × -6 = -30.

Find out the prime factors of 30. i.e. 30 = 2 × 3 × 5

Here, we cannot find a pair of values that give 5x on addition or subtraction. Hence, the method of factorization will not work here and is limited.

We can solve this equation either by using the quadratic formula or method of completing squares.

Method of Completing the Squares

We know that,

(p + q)2 = p2 + 2pq + q2

(p – q)2 = p2 – 2pq + q2

We can see that (p + q)2 and (p – q)2 expands in the general form of a quadratic equation if we substitute a = x. Thus, we can find the solution of the quadratic equation by converting the equation from general form to (x + q)2 = d or (x – q)2 = d form and thus the roots can be found.

Let us solve an example problem to understand this method.

Example: Find the roots of a quadratic equation given below.

5x2 + 5x – 6 = 0

We have to write the equation in the form x2 + 2qx + q2 = d

First dividing the equation by 5 so that coefficient of x2 becomes 1.

x2 + x – 6/5 = 0

Adding 6/5 to both sides. We get,

x2 + x = 6/5

Transform the equation in such a way so that the coefficient of x is in the form 2qx.

x2 + 2(1/2)x = 6/5

The value of q = 1/2. Adding q2 on both sides.

x2 +  2(1/2)x + 1/4 = 6/5 + 1/4

We can see that x2 +  2(1/2)x + 1/4 is in the form of x2 + 2qx + q2 and we can apply the formula (p + q)2 = p2 + 2pq + q2.

(x + 1/2)2 = 29 / 20

Taking the square roots on both sides.

x + 1/2 = ± √29 / 20

We have two answers,

x + 1/2 = √29 / 20 and x + 1/2 = -√29 / 20

x  = 0.704 and x = -1.704

Thus, the roots of the equation are 0.704 and -1.704.