Quadratic Equation is a second-degree polynomial equation with one variable x. The graph of quadratic equations makes nice curves. For every equation, we have two values of the variables called the roots. Get the formula and simple step by step process to solve the roots of any quadratic equation in the following sections.

The quadratic equation is an equation where the highest exponent of the variable is square. The standard form of quadratic equations is ax² + bx + c = 0. Where x is the variable, a, b, c are the constants and a should not be equal to zero. The power of x should not be negative and must be 2.

The formulas to find the solution or roots of the quadratic equation are given below:

(α, β) = [-b ± √(b² – 4ac)] / 2a

The values of the variable satisfying the given quadratic equation are called its roots. The roots of the quadratic equation details are mentioned here.

• Both roots of the equation are zero when b and c are zero.
• One of the roots of the quadratic equation is zero and the other is -b/a if c = 0
• The two roots are reciprocal to each other when the coefficients a and c are equal.

### Nature of Roots of Quadratic Equation

In the quadratic equation formula, the term (b² – 4ac) is called the discriminant of the equation and it gives information about the nature of the roots. Below listed are the conditions that define the nature of roots.

• If the discriminant value is zero, then the equation will have equal roots i.e α = β = -b/2a.
• If the discriminant value is greater than zero, then the equation will have real roots.
• If the discriminant value is less than zero, then the equation will have imaginary roots.
• If the discriminant is greater than zero and it is a perfect square, then the equation will have rational roots.
• If the discriminant is greater than zero and it is not a perfect square, then the equation will have irrational roots.
• If the discriminant is greater than zero, perfect square, and a = 1, b, c are integers, then the equation will have integral roots.

### How to Solve Quadratic Equations?

By solving the quadratic equations, you will get two roots that satisfy the equation. The easy steps to solve the quadratic equation roots are given here. The two ways to find the quadratic equations roots are the algebraic method and the graphical method. Here, we will learn about those methods.

Algebraic Method:

• Convert the given equation in the form of ax² + bx + c = 0, get a, b, c values.
• Substitute the values in the quadratic formula.
• Perform all the required calculations and find the roots.
• Another simple way is by factorizing the quadratic equation.
• The obtained factors are the roots.

Graphical Method:

• Plot the graph for the random values of x.
• The points where the curve meets the x-axis are the roots.

(i) 5x² + 3x + 2 = 0 is a quadratic equation.

(ii) x – (1/x) = 6 is a quadratic equation

On solving this, we get x * x – 1 = 6 * x

x² – 1 = 6x

x² – 6x – 1 = 0

(iii) x² + √2x + 7 = 0 is not a quadratic equation.

(iv) x² + 5 = 0 is a quadratic equation.

(v) x² = 0,√ x² + 2x + 1 are quadratic equations.

### Solved Examples on Finding Roots of the Equation

Example 1.

Solve the roots of equation 3x² – 22x – 16 by factorization?

Solution:

Given the quadratic equation is 3x² – 22x – 16

= 3x² – 24x + 2x – 16

= 3x(x – 8) + 2(x – 8)

= (x – 8) (3x + 2)

x – 8 = 0 and 3x + 2 = 0

x = 8 and 3x = -2, x = -2/3

The roots of the equation are α = 8, β = -2/3.

Therefore, solution set (8, -2/3)

Example 2.

Solve 9x² + 25x + 10 = 0 by using the quadratic formula.

Solution:

Given that,

9x² + 25x + 10 = 0

Roots of equation formula is (α, β) = [-b ± √(b² – 4ac)] / 2a

In the equation, a = 9, b = 25, and c = 10

Substitute these values in the equation.

α = [-b + √(b² – 4ac)] / 2a

= [-9 + √(9² – 4 (9) (10)] / 2 (9)

= [-9 + √(81 – 360)] / 18

= [-9 + √(-279)] / 18

= [-9 + √(279i²)] / 18

= [-9 + 16.70i] / 18

β = [-b – √(b² – 4ac)] / 2a

= [-9 – √(9² – 4 (9) (10)] / 2 (9)

= [-9 – √(81 – 360)] / 18

= [-9 – √(-279)] / 18

= [-9 – √(279i²)] / 18

= [-9 – 16.70i] / 18

Roots are α = [-9 + 16.70i] / 18, β = [-9 – 16.70i] / 18

Therefore, solution set ([-9 + 16.70i] / 18, [-9 – 16.70i] / 18)

Example 3.

Find the roots of the equation 1/(x + 4) – 1/(x – 7) = 11/30.

Solution:

Given that,

1/(x + 4) – 1/(x – 7) = 11/30

[(x – 7) – (x + 4)] / (x + 4) (x – 7) = 11/30

[x – 7 -x – 4] / (x² + 4x – 7x – 28) = 11/30

(-11) / (x² – 3x – 28) = 11/30

-1/ (x² – 3x – 28) = 1/30

-30 = x² – 3x – 28

x² – 3x – 28 + 30 = 0

x² – 3x + 2 = 0

x² – 2x – x + 2 = 0

x(x – 2) -1(x – 2) = 0

(x – 1) (x – 2) = 0

(x – 1) = 0 and (x – 2) = 0

x = 1, x = 2

The roots of quadratic equation is α = 1, β = 2.

Therefore, the solution set is (1, 2)

Example 4.

Find the values of a for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.

Solution:

Given that,

(x – a) (x – 10) + 1 = 0

x² – ax – 10x + 10a + 1 = 0

x² – (a + 10)x + 10a + 1 = 0

Discriminant = b² – 4ac = (a + 10)² – 4 . 1. (10a + 1)

= a² + 20a + 100 – 40a – 4

= a² – 20a + 100 – 4

= (a – 10)² – 4

The quadratic equation will have integral roots, if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.

(a – 10)² – Discriminant = 4

Since discriminant is a perfect square. Hence, the difference of two perfect square in R.H.S will be 4 only when D = 0 and (a – 10)² = 4.

(a – 10) = ± 2

a = 2 + 10 or a = -2 + 10

= 12 and 8

Therefore, values of a are 8, 12.

1. What are the uses of the quadratic equations?

The quadratic equations can be used in calculating the area of rooms, speed of an object, determining product profit in the business. It can also be used in athletics and sports.