Roots of Quadratic Equation

The roots of a quadratic equation always satisfy the value of a variable. They are also known as quadratic equation solutions or zeros. Graphically, the root of a quadratic equation represents the x-intercepts of its graph, so any quadratic equation of degree 2 has a maximum of 2 roots.  

Now let’s see how to find the roots of a quadratic equation. For a general quadratic equation ax² + bx + c = 0, where a ≠ 0, let’s find the root using completing the square.

ax² + bx + c = 0

ax² + bx = -c

Dividing the equation by a, we get: 

(a/a)x² + (b/a)x = -c/a

x² + (b/a)x = -c/a

Taking half of the coefficient of x

Now the coefficient of x is b/a, half of b/a = b/(2a)

Squaring and adding it to both sides: that is, b²/(2a)² = b²/4a²
Adding it to both sides: 

x² + (b/a)x + b²/4a² =b²/4a² -ca 

Using (a + b)² formula, (x + (b/2a))² = (b² - 4ac)/ 4a²

Taking the square root on both sides, 

x+ (b/2a) = ±√(b2 - 4ac/2a)

Subtracting b/(2a) on both sides

x+ (b/2a) - (b/2a) = ±√(b2 - 4ac/2a) - (b/2a)

x = -b ± √b² - 4ac/2a

Therefore, this is the quadratic formula used to find the roots of a quadratic equation.

Solving the quadratic equation is the process of finding the roots of a quadratic equation. We know the equation to find the roots of a quadratic equation. Apart from that, there are different methods to find the roots, such as: 

• Factoring
   
• Quadratic formula
   
• Completing the square
   
• Graphing

Let’s learn these methods in detail using an example: 2x² + 5x - 3 = 0.

To find the roots of a quadratic equation using factoring, we first find the factors of the equation on the left-hand side, then set each factor equal to zero, and then solve the equation to find the value of x. 

Let's learn how to find the root of a quadratic equation with the example, 2x² + 5x - 3 = 0

Factoring the quadratic equation: 2x² + 5x - 3 

To factor, find two numbers whose product is -6 and sum is 5.

Here, the numbers are 6 and -1 (6 × -1 = -6 and 6 + (-1) = 5)

Rewriting the equation: 2x² + 6x - x - 3

Grouping the terms: (2x² + 6x) - (x + 3)

Factoring: 2x(x + 3) - 1(x + 3)

= (2x - 1)(x + 3)

Setting each factor equal to zero: 

2x - 1 = 0 or x + 3 = 0

Solving the equations to find the value of x, 

2x - 1 = 0 

2x = 1

x = ½

x + 3 = 0

So, x = -3

So, the roots of the equation are: x = ½ and x = -3.

The quadratic formula is a standard and commonly used method to find the roots of a quadratic equation. For any quadratic equation in the form: ax² + bx + c = 0, the quadratic formula is x = -b ± √b² - 4ac/2a. So to find the root value, we first identify the values a, b, and c and substitute them in the formula. 

For example, to find the value of x in 2x² + 5x - 3 = 0. 

Here, a = 2, b = 5, and c = -3

Substituting it in, x = -b ± √b² - 4ac/2a

x = (-5 ± (√5² - (4 × 2 × -3))) / 2 ×2

x = (-5 ± (√25 - (-24))) / 4

x = (-5 ± √49) / 4

x = (-5 ± 7) / 4

That is x = (-5 + 7) / 4 and x = (-5 - 7) / 4

x = (-5 + 7) / 4 = 2/4 = ½

x = (-5 - 7) / 4 = -12/4 = -3

Therefore, the value of x is ½ and -3. 

In completing the square method, we convert the left-hand side of the quadratic equation into a perfect square and then solve it by taking the square roots. This method applies to all types of quadratic equations, including real and complex numbers. 

For example, finding the roots of the quadratic equation, 2x² + 5x - 3 = 0
2x² + 5x - 3 = 0

Moving the constant to the right-hand side: 

2x² + 5x = 3

Dividing the equation by a coefficient of x2, here it is 2

So, (2x² / 2) + (5x / 2) = 3/2

x² + (5/2)x = 3/2

Taking half of the coefficient of x, that is 5/2 ÷ 2 = 5/4

Squaring it: (5/4)² = 25/16

Adding it to both sides:
 
x² + (5/2)x + 25/16 = 3/2 + 25/16

So, x² + (5/2)x + 25/16 = 3/2 + 25/16, becomes 

(x + 5/4)² = 49/16

Taking the square root on both sides:

x + 5/4 = ±7/4

So, the value of x can be:

x + 5/4 = 7/4 ⇒ x = 7/4 - 5/4 = 2/4 = ½

x + 5/4 = -7/4 ⇒ x = -7/4 - 5/4 = -12/4 = -3

Thus, the value of x can be ½ and -3.

The graphing method involves plotting the equation as a parabola on the coordinate plane and finding where the graph intersects the x-axis. In the graph, the point where the graph intersects the x-axis is the root of the quadratic equation. Let’s learn this in detail using an example, 2x² + 5x - 3 = 0. 

The quadratic equation is written as a function: 

f(x) = 2x² + 5x -3 

As the intersecting points are in -3 and ½, so the roots of the quadratic equation are x = -3 and x = ½.

The nature of the roots of a quadratic equation helps us to understand how many roots exist and what kind of roots they are, that is, whether they are real or complex. The roots can be real and distinct, complex, or real and equal. The nature of the roots can be determined using the discriminant; the discriminant of the quadratic equation ax² + bx + c = 0 is D = b² - 4ac. 

The formula of quadratic equation: x = (-b ± (√b² - 4ac)) / 2a

As D = b² - 4ac

Then x = (-b ± √D) / 2a

The value of D determines the nature of the roots, that is, whether D is positive, negative, or zero. 

Nature of Roots When D > 0

If D > 0, the above formula becomes, x = (-b ± (√positive number)) / 2a. If the discriminant is positive, that is b² - 4ac > 0, because √a positive number is real and non-zero. So, when D > 0, the quadratic formula gives two real and different roots.  

Nature of Roots When D < 0

The quadratic formula, x = (-b ± (√b² - 4ac)) / 2a becomes, x = (-b ± (√negative number)) / 2a, when D < 0. The square root of a negative number is an imaginary number, so the equation has two complex conjugate roots. 

Nature of Roots When D = 0

If the discriminant is zero, that is, b² - 4ac = 0, it means the square root is 0. So here the formula x = (-b ± (√0)) / 2a = -b/2a. 

To understand how two solutions are related to the coefficients of x we use the sum of the roots of the quadratic equation 

For a general quadratic equation: ax2 + bx + c = 0, the quadratic formula to find the root is x = (-b ± (√b2 - 4ac)) / 2a
Consider the roots as: x1 and x2 so, 
x1 = (-b + (√b2 - 4ac)) / 2a, x2 = (-b - (√b2 - 4ac)) / 2a

Adding x₁ + x₂: 

x₁ + x₂ = (-b + (√b² - 4ac)) / 2a + (-b - (√b² - 4ac)) / 2a

= -b/2a - b/2a

= -2b/2a

= -b/a

Therefore, the sum of the quadratic equation ax² + bx + c = 0 is -b/a. 

In this section, let’s find how the product of roots of a quadratic equation is related to the constant and the value of a. The product of the roots of the quadratic equation ax² + bx +  c = 0 can be x₁ and x₂.

x₁ = x₁ = (-b + (√b² - 4ac)) / 2a, x₂ = (-b - (√b² - 4ac)) / 2a

x₁ × x₂ = (-b + (√b² - 4ac)) / 2a × (-b - (√b² - 4ac)) / 2a

By using the identity a² - b² = (a + b) (a - b)

x₁ × x₂ = (-b)² - (√b² -4ac)² / (2a)²

= b² - (b² - 4ac) / 4a²

x₁ × x₂ = 4ac / 4a²

= c/a 

Therefore, the product of the roots of a quadratic equation is x₁x₂ = c/a.

The quadratic equations and their roots are used in fields like physics, engineering, and economics. Here are some applications of the roots of quadratic equations

• In engineering, the quadratic equation is used to design the structure and suspension of bridges. 

• To find the optimal prices and quantities for a product to maximize profits, we use quadratic equations. For example, if the profit p(x) depends on the number of units sold, where p(x) = -5x² + 200x - 1500, we use the quadratic formula to find the profit. 

• In various optimization problems, we use quadratic equations to find the minimum and maximum values of functions. 

• In sports, to calculate and predict the trajectory of the projectile and its range. In basketball, to predict the trajectory of the ball when a player takes a shot follows a parabolic path.