Roots of Quadratic Equation

Roots of Quadratic Equations are the values of the variable for which the quadratic equation gets satisfied. The roots of a quadratic equation are also called zeros of a quadratic equation.

A quadratic equation is an equation in which the maximum power of a variable is 2. Hence, there can be a maximum of two possible zeros or roots or solutions of a quadratic equation.

Let’s learn more about Roots of Quadratic equations with formulas and examples below.

The roots of a quadratic equation, which is typically written as ax² + bx + c = 0 where a, b, and c are constants and a ≠ 0.

Roots of a Quadratic Equation are the values of the variable let’s say x for which the equation gets satisfied.

• Roots of Quadratic Equations are also called Zeros of a Quadratic Equation or Solutions of a Quadratic Equation.

• Quadratic equations are mathematical expressions of the form ax² + bx + c = 0, where a, b, and c are constants, and x represents the variable.

• Solving for the values of x that make this equation true yields the roots of the quadratic equation.

Methods to Find Root of Quadratic Equation

The methods to find Roots or Zeroes of a Quadratic Equation are :

• Factoring
• Quadratic Formula
• Completing the Square
• Graphical Method

To Find the roots of Quadratic Equations follow the methods mentioned below:

Roots of Quadratic Equation Formula

x = [-b±√(b² – 4ac)]/2a

Example: The length of sides of a rectangle is given by x – 3 and x – 5 and the area of the rectangle is 3 unit². Find the sides of the rectangle.

Solution:

Area of rectangle = length × breadth = (x – 3)(x – 5) = 3

Area = x² – 8x + 15 = 3

= x² – 8x + 12 = 0

Discriminant = b² – 4ac = 64 – (4(1)(12)) = 64 – 48 = 16

 x = [-b ± √(b²-4ac)]/2a = [-(-8) ± √16]/2 = (8±4)/2

 x = 12/2 or 4/2

x = 2 or 6

When x is 2, sides are x – 3 = 2 – 3 = -1 and x – 5 = 2 – 5 = -3.

Since the length of sides cannot be equal therefore x = 2 is not a valid answer.

When x is 6, sides are x – 3 = 6 – 3 = 3 and x – 5= 6 – 5 =1.

Therefore, x = 6 is the valid answer and the sides are 3 and 1.

Roots of Quadratic Equation by Factoring

A quadratic equation can be considered a factor of two terms. Like ax² + bx + c = 0 can be written as (x – x₁)(x – x₂) = 0 where x₁ and x₂ are roots of quadratic equation.

Step 1: Find two numbers such that their product = ac and their sum = b.

Step 2: Write x coefficient as the sum of these two numbers and split them such that you get two terms for x.

Step 3: Factor the first two as a group and the last two terms as a group.

Step 4: Take common factors from these and on equating the two expressions with zero after taking common factors and rearranging the equation we get the roots.

Example: Find the roots of the quadratic equation x² + 3x = 18

Solution:

x² + 3x – 18 = 0

Step 1: 6 and -3 are the numbers whose sum is equal to b and the product is equal to a.c.

Step 2: x² + (6-3)x – 18 = x² + 6x -3x – 18 =0  

Step 3: x(x + 6) – x(x + 6) = 0

Step 4: Take (x + 6) as common. 

(x + 6)(x – 3) = 0

x = -6 or x = 3

In a factorizing method, you don’t need to always find these two numbers easily(especially in the case when roots are imaginary or irrational) so it is better to use the quadratic formula.

The sum of the roots of a quadratic equation of the form ??²+??+?=0 can be derived directly from the coefficients of the equation. The sum of the roots is denoted by ?+? is given by:

α + β = -b/a

This relationship comes from Vieta’s formulas, which relate the coefficients of a polynomial to sums and products of its roots. In the case of a quadratic equation:

• Sum of the roots α + β is equal to -b/a.
• Product of roots α.β is equal to c/a​.

Nature of Roots of Quadratic Equation depends on the discriminant of the quadratic equation. The discriminant of a quadratic equation is given by b² – 4ac.  It is so because in the quadratic formula square root of the discriminant is there.

Case 1: If b² – 4ac > 0 roots are real and different. As the discriminant is greater than 0 then the square root of it will real and unequal.

• If b² – 4ac is a perfect square then roots are rational. As the discriminant is a perfect square, so we will have an integer as a square root of the discriminant. Hence, the roots are rational numbers.

Example: Let quadratic equation be x²– 5x + 6 = 0.

Then the discriminant of the given equation is b² – 4ac = (-5)² – 4×1×6 = 25-24 = 1

According to Shridharacharaya formula

x = [-b±√(b²-4ac)]/2a = x = [-(-5) ± √1]/2 

x₁ = [-(-5) + √1]/2 = 6/2 = 3

x₂ = [-(-5) – √1]/2 = 4/2 = 2

Therefore, the roots are 3,2. Both are rational and different.

• If b² – 4ac is not a perfect square then the square root of the discriminant is irrational hence roots are irrational and occur in pairs.

Example: Let quadratic equation be x²-7x+8 = 0.

Then the discriminant of the given equation is

b² – 4ac = (-7)² – 4×1×8 = 49-32 = 17

According to Shridharacharaya formula

x = [-b±√(b²-4ac)]/2a =  x = [-(-7) ± √17]/2 

x₁ = [-(-7) + √17]/2 = [7 + √17]/2

x₂ = [-(-7) – √17]/2 = [7 – √17]/2

Therefore, the roots are [7 + √17]/2,[7 – √17]/2. Both are irrational and in pairs.

Case 2: If b² – 4ac = 0 roots are real and equal.

Example: Let the quadratic equation be 3x²-6x+3=0.

Then the discriminant of the given equation is

b² – 4ac=(-6)² – 4×3×3 = 36 – 36 = 0

According to Shridharacharaya formula

x = [-b±√(b²-4ac)]/2a =  x = [-(-6) ± √0]/[(2)(3)]

x₁ = [-(-6) + √0]/2 = 6/6 = 1

x₂ = [-(-6) – √0]/2 = 6/6 = 1

Therefore, the roots are 1,1. Both are real and equal.

Case 3: If b² – 4ac < 0 roots are imaginary, or you can say complex roots. It is imaginary because the term under the square root is negative. These complex roots will always occur in pairs i.e., both the roots are conjugate of each other.

Example: Let the quadratic equation be x²+6x+11=0.

Then the discriminant of the given equation is

b² – 4ac=(6)² – 4×1×11 = 36-44 = -8

According to Shridharacharaya formula

x = [-b±√(b²-4ac)]/2a = x = [-(6) ± √(-8)]/2 

x₁ = [-(6) + √(-8)]/2 = [-6 + √8i]/2 = 2[-3 + √2i]/2 = -3 + √2i

x₂ = [-(6) – √(-8)]/2 = [-6 – √8i]/2 = 2[-3 – √2i]/2 = -3 – √2i

Therefore, the roots are 3,2. Both are imaginary and conjugate of each other(in pairs).

Maximum/minimum value of the quadratic function is found at x = -b/2a

Proof: 

We get maxima or minima when d(f(x))/dx = 0

On differentiating quadratic function f(x) = ax² + bx + c

We get,

2ax + b = 0 

x = -b/2a

This x is either maxima(a < 0) or minima(a > 0). 

1. When D > 0

In the case of D > 0, the graph of the quadratic equation ax² + bx + c = 0 touches the x-axis at two distinct points which indicates two possible values of x i.e. real and unequal roots. Graph of b²– 4ac > 0 is in the image shown below:

2. When D = 0

In the case of D = 0, the graph of the quadratic equation ax² + bx + c = 0 touches the x-axis at only one point which indicates one unique value of x i.e. real and equal roots. Graph of b²– 4ac = 0 is shown in the image added below:

3. When D < 0

In the case of D < 0, the graph of the quadratic equation ax² + bx + c = 0 doesn’t touch the x-axis which indicates no possible values of x i.e. no real roots. Graph of b²– 4ac < 0 is shown in the image below:

Must Read,

• Solving Quadratic Equations
• Quadratic Formula

Example 1. Height of a triangle is less than 4 cm than the base. The area of triangle is 30 cm². Find the height and base of the triangle.

Solution:

Let the base of triangle be x cm then height is x-4 cm 

Area of triangle = 1/2*height*base = 1/2*(x)(x – 4)=30

Area = x² – 4x  = 30*2

        = x² – 4x = 60

        = x² – 4x – 60 = 0

Discriminant = (-4)² – 4(1)(-60) = 16+240 = 256

x = [-b±√(b² – 4ac)]/2a = [-(-4)±√256]/2 = (4±16)/2

x = 20/2 or -12/2

x = 10 or -6

As side cannot be negative, therefore -6 is not correct.

So, when x is 10, base =10 cm and height =x – 4 = 10 – 4 = 6 cm

Therefore, x = 10 is the valid answer and the base and height of the triangle are 10 and 6 respectively.

Example 2. Volume of a box is 600 inches². The length of the box is 2 inches less than the width. The height of the box is 5 inches. Find the dimensions of the box.

Solution:

Let the width of box be x inches then length = x – 2 inches.

Volume of box =Length* Width *Height = (x-2)(x)5= 600

x² – 2x = 120 => x² – 2x -120 = 0

x² + 10x – 12x – 120 = 0

x(x + 10) – 12(x + 10)=0

(x – 12)(x + 10)=0

x = 12 or x = -10

As width can not be negative, therefore, -10 is not correct.

When x = 12, width = 12 inches, length = x – 2 = 12 – 2 = 10 inches, height = 5 inches 

Example 3. A ball is thrown from the top of a building. its height in meters above the ground as a function of time is given by h(t) = -4t² + 24t + 3. a) How much time does it take to reach the maximum height and what is the maximum height? b) Find also the time at which the ball hits the ground.

Solution:

a) Since a<0, therefore the time to reach maximum height is = -b/2a.

t = -24/(2(-4)) = -24/-8

t = 3 sec.

Height = h(t) = -4(3)² +24(3)+3 = 39 meters.

b) When the ball hits the ground h(t) = 0.

-4t² + 24t + 3 = 0

The discriminant of the given equation is

b² – 4ac=(24)² – 4×(-4)×3 = 576 + 48 = 624

According to Quadratic Formula

x = [-b±√(b²-4ac)]/2a = x = [-(24) ± √624]/[(2)(-4)]

x₁ = [-24 + √624]/-8= 4 [-6 + √39]/-8 = [-6 + √39]/-2 = [6 – √39]/2 = -0.122499 = -0.1225(approx)

x₂ = [-24 – √624]/-8= 4 [-6 – √39]/-8 = [-6 – √39]/-2 = [6 + √39]/2 = 6.122499 = 6.1225(approx)

As time can not be negative, therefore, [6 – √39]/2 sec is not correct.

Therefore, the ball hits the ground at [6 + √39]/2 = 6.122499 = 6.1225 sec

Problem 1: Find the roots of the following Quadratic Equations:

• 3x² – 7x + 2 = 0
• x² + 4x + 4 = 0
• 2x² + 5x – 3 = 0
• 4x² – 12x + 9 = 0
• x² – 6x + 9 = 0

Problem 2: The sum of two consecutive integers is 31. Find the two integers.

Problem 3: A quadratic equation of the form ax² + bx + c = 0 has roots x = 3 and x = -2. Find the values of a, b and c.

Problem 4: A ball is thrown into the air from a height of 5 feet with an initial velocity of 40 feet per second. The height h of the ball above the ground after t seconds can be modelled by the quadratic equation h(t) = -16t² + 40t + 5. How long does it take for the ball to reach its maximum height?

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of the second degree, typically written in the form ax² + bx + c = 0, where a, b, and c are constants, and x is the variable.

What are the Roots of a Quadratic Equation?

The roots of a quadratic equation are the values of ‘x’ that satisfy the equation, making it equal to zero. These are the values where the graph of the quadratic function intersects the x-axis.

How do you find the Roots of Quadratic Equation?

To find the roots of a quadratic equation using the quadratic formula, i.e., x = (-b ± √(b² – 4ac)) / (2a).

How many Roots Can a Quadratic Equation Have?

A quadratic equation can have either two real roots, one real root (a repeated root), or two complex roots (conjugate pair of complex numbers).

Can a Quadratic Equation have no Real Roots?

Yes, if the discriminant is negative (b² – 4ac < 0), the quadratic equation will have two complex roots and no real roots.

How do you Determine the Nature of the Roots of a Quadratic Equation?

You can determine the nature of the roots by calculating the discriminant (b² – 4ac) and examining its value in relation to zero.

• If discriminant > 0, two real roots,
• If discriminant = 0, one real repeated root, and
• If discriminant < 0, two complex roots i.e., no real roots.

What is the Sum and Product of the Roots of Quadratic Equation?

For a quadratic equation in the form ax² + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a.

What are Four Types of Roots of Quadratic Equations?

Four types of roots in quadratic equations are:

• Two Real and Distinct Roots
• Two Real and Equal Roots
• One Real and Repeated Root
• Two Complex Roots

Why are Alpha (α) and Beta (β) used to Represent the Roots of Quadratic equations?

Alpha (α) and beta (β) are commonly used to represent the roots of a quadratic equation for simplicity and consistency in mathematical notation.