An equation of the form ax2 + bx + c = 0, where a, b and c are real numbers and a 0 is called a quadratic equation in variable x. This is also known as the standard form of the quadratic equation. For example, x2 + 5x + 6 = 0, 2x2 + 5x – 8 = 0 and 4x2 – 11x + 7 are the quadratic equations.
In fact, an equation of the form p(x) = 0, where p(x) is a polynomial of degree 2 is a quadratic equation.

Let us consider the quadratic equation, x2 + 5x – 6 = 0.
Now, when we substitute x = 1, we get (1)2 + 5 × 1 – 6 = 0
Thus, 0 = 0 means LHS = RHS
So, 1 is the solution or root of the above quadratic equation.
In general, a real number a is called the root or solution of the quadratic equation ax2 + bx + c = 0, a 0 if it satisfies the quadratic equation, i.e., aa2 + ba + c = 0.

## Solution of Quadratic Equations by Factorization

We know that, x2 – 7x + 10 = 0 can also be written as x2 – 2x – 5x + 10 = 0
Or, x (x – 2) – 5 (x – 2) = 0 or (x – 2) (x – 5) = 0 (using factorization)
Now, using zero-product rule which states that if a and b are two numbers or expressions and if ab = 0 then either a = 0 or b = 0 or both a = 0 and b = 0.
We have, (x – 2) = 0 or (x – 5) = 0
Thus, x = 2 or x = 5. Hence, x = 2 and x = 5 are the solutions of the equation. Thus, in order to solve the quadratic equation using factorization, we proceed as follows:
1. Express the given equation in the form ax2 + bx + c = 0.
2. Factorize left hand side and equate each factor to zero.
3. Solve each resultant in a linear equation.

Example 1: Solve: x2 + 2x – 3 = 0
Solution: Given, x2 + 2x – 3 = 0 implies x2 + 3x – x – 3 = 0
x (x + 3) – 1 (x + 3) = 0 implies (x – 1) (x + 3) = 0
x – 1 = 0 or x + 3 = 0
Thus, x = 1 or x = –3
Verification: When x = 1, LHS = 12 + 2 × 1 – 3 = 1 + 2 – 3 = 0 = RHS
When x = –3, LHS = (–3)2 + 2 × (–3) – 3 = 9 – 6 – 3 = 0 = RHS
Thus, x = 1 and –3 are the roots of the given quadratic equation.

Example 2: Solve: 2x2 – 7x + 3 = 0
Solution: Given, 2x2 – 7x + 3 = 0
2x2 – 6x – x + 3 = 0
2x (x – 3) – 1 (x – 3) = 0
(2x – 1) (x – 3) = 0
2x – 1 = 0 or x – 3 = 0
Thus, x = 1/2 or x = 3
Hence, x = 1/2 and 3 are the roots of the equation.

Example 3: Solve for x: 4x2 + 12x + 9 = 0
Solution: Given, 4x2 + 12x + 9 = 0
4x2 + 6x + 6x + 9 = 0
2x (2x + 3) + 3 (2x + 3) = 0
(2x + 3) (2x + 3) = 0
2x + 3 = 0 or 2x + 3 = 0
x = -3/2 or x = -3/2
Thus, the roots of the equation are -3/2 , -3/2 .

If b2 – 4ac ≥ 0, then the roots of the quadratic equation ax2 + bx + c = 0 are given by
This formula for finding the roots of a quadratic equation is known as the quadratic formula.
Since this formula was given by Sreegharacharya, this is also called Sreedharacharya formula.

Example 4: Solve 3x2 - 5x + 2 = 0 using the quadratic formula.
Solution: The equation is 3x2 + 5x + 2 = 0.
Here, a = 3, b = -5 and c = 2
So, b2 – 4ac = (-5)2 – 4 × 3 × 2 = 25 – 24 = 1 ˃ 0.
Therefore, x = (-b ± √b2 – 4ac)/2a
x = (5 ± √1)/2 × 3 = (5 ± 1)/6
x =  (5 + 1)/6 and (5 - 1)/6
x = 1 and 2/3
So, the roots are 1 and 2/3.

Example 5: The sum of the number and its reciprocal is given by 10/3 . Find the number.
Solution: Let x be the given number then its reciprocal is 1/x . Given, x + 1/x = 10/3
x + 1/x – 10/3 = 0
3x × x + 3 – 10 × x = 0
3x2 + 3 – 10x = 0
3x2 – 10x + 3 = 0
3x2 – 9x – x + 3 = 0
3x (x – 3) – 1 (x – 3) = 0
(3x – 1) (x – 3) = 0
3x – 1 = 0 or x – 3 = 0
x = 1/3 or x = 3
Thus, the given number is 3.

Example 6: The perimeter of a rectangle is 46 cm and its area is 120 cm2 . Find the dimensions of the rectangle.
Solution: Let x and y be the length and breadth of the rectangle respectively.
Perimeter of the rectangle = 2 (length + breadth) = 2 (x + y)
46 = 2 (x + y)
x + y = 23
y = 23 – x
Now, area of the rectangle = length × breadth = xy
120 = x (23 – x)       (putting the value of y)
120 = 23x – x2
x2 – 23x + 120 = 0
x2 – 15x – 8x + 120 = 0
x (x – 15) – 8 (x – 15) = 0
(x – 8) (x – 15) = 0
x – 8 = 0 or x – 15 = 0
x = 8 or x = 15
When length (x) = 8 cm, breadth (23 – x) = 15 cm or when length (x) = 15 cm, breadth (23 – x) = 8 cm. Hence, the dimensions of the rectangle are 15 cm by 8 cm.

## Discriminant (D) of a Quadratic Equation

In a quadratic equation ax2 + bx + c = 0, b2 – 4ac is called the discriminant (D).
Thus, the discriminant (D) = b2 – 4ac
To find the nature of the two roots of a quadratic equation, we first find the discriminant (D) and then compare the value of D with 0.
1.      If D > 0, the two roots of the quadratic equation are real and distinct.
2.      If D = 0, the two roots of the quadratic equation are real and equal.
3.      If D < 0, the two roots of the quadratic equation are imaginary.

Example 7: Find the discriminant (D) of the quadratic equation 2x2 + x – 6 = 0, and hence find the nature of its roots.
Solution: The given equation is 2x2 + x – 6 = 0.
Here, a = 2, b = 1 and c = -6.
Discriminant (D) = b2 – 4ac = 12 – 4 × 2 × -6 = 1 + 48 = 49 > 0
Here, D > 0, hence the roots of the equation are real and distinct.