**Quadratic
Equations **

An equation of the form ax

^{2}+ 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, x^{2}+ 5x + 6 = 0, 2x^{2}+ 5x – 8 = 0 and 4x^{2}– 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.

**Solution
of Quadratic Equations**

Let us
consider the quadratic equation, x

^{2}+ 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 ax

^{2}+ bx + c = 0, a ≠ 0 if it satisfies the quadratic equation, i.e., aa^{2}+ ba + c = 0.**Solution of Quadratic
Equations by Factorization**

We know
that, x

^{2}– 7x + 10 = 0 can also be written as x^{2}– 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 ax

^{2}+ 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: x

^{2}+ 2x – 3 = 0

**Solution:**Given, x

^{2}+ 2x – 3 = 0 implies x

^{2}+ 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 = 1

^{2}+ 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: 2x

^{2}– 7x + 3 = 0

**Solution:**Given, 2x

^{2}– 7x + 3 = 0

2x

^{2}– 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: 4x

^{2}+ 12x + 9 = 0

**Solution:**Given, 4x

^{2}+ 12x + 9 = 0

4x

^{2}+ 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 .

**Solution
of the Quadratic Equation using Quadratic Formula**

If b

^{2}– 4ac ≥ 0, then the roots of the quadratic equation ax^{2}+ bx + c = 0 are given by
Since this formula was given by Sreegharacharya, this is also called

**Sreedharacharya formula**.**Example 4:**Solve 3x

^{2}- 5x + 2 = 0 using the quadratic formula.

**Solution:**The equation is 3x

^{2}+ 5x + 2 = 0.

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

So, b

^{2}– 4ac = (-5)^{2}– 4 × 3 × 2 = 25 – 24 = 1 ˃ 0.
Therefore, x = (-b ± √b

^{2}– 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

3x

^{2}+ 3 – 10x = 0
3x

^{2}– 10x + 3 = 0
3x

^{2}– 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 cm

^{2}. 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 –
x

^{2}
x

^{2}– 23x + 120 = 0
x

^{2}– 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.

**Nature of
Roots of a Quadratic Equation**

**Discriminant
(D) of ****a Quadratic Equation**** **

In a
quadratic equation ax

^{2}+ bx + c = 0, b^{2}– 4ac is called the**discriminant (D)**.
Thus, the
discriminant

**(D) = b**^{2}– 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 2x

^{2}+ x – 6 = 0, and hence find the nature of its roots.

**Solution:**The given equation is

**2x**

^{2}+ x – 6 = 0.

Here, a = 2,
b = 1 and c = -6.

Discriminant
(D) = b

^{2}– 4ac = 1^{2}– 4 × 2 × -6 = 1 + 48 = 49 > 0
Here, D >
0, hence the roots of the equation are real and distinct.