# NATURE OF THE ROOTS OF A QUADRATIC EQUATION

"Nature of the roots of a quadratic equation" is the stuff which is required to the students who study math in school level.

To understand the nature of the roots of a quadratic equation, let us consider the general form a quadratic equation.

ax² + bx + c = 0

(Here a, b and c are real and rational numbers)

To know the nature of the roots of a quadratic-equation, we will be using the discriminant  "b² - 4ac".

Because "b² - 4ac"discriminates the nature of the roots.

Let us see how this discriminant  "b² - 4ac" can be used to know the nature of the roots of a quadratic-equation.

## Examples

Example 1 :

Examine the nature of the roots of the following quadratic equation.

x² + 5x + 6 =0

Solution :

If x² + 5x + 6 =0 is compared to the general form ax² + bx + c =0,

we get a = 1, b = 5 and c = 6.

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = 5² - 4(1)(6)

b² - 4ac = 25 - 24

b² - 4ac = 1 (>0 and also a perfect square)

Hence, the roots are real, distinct and rational.

Let us look at the next example on nature of the roots of a quadratic equation.

Example 2 :

Examine the nature of the roots of the following quadratic equation.

2x² - 3x + 1 =0

Solution :

If 2x² - 3x + 1 =0 is compared to the general form ax² + bx + c =0,

we get a = 2, b = -3 and c = 1.

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = (-3)² - 4(2)(-1)

b² - 4ac = 9 + 8

b² - 4ac = 17 (>0 but not a perfect square)

Hence, the roots are real, distinct and irrational.

Let us look at the next example on nature of the roots of a quadratic equation.

Example 3 :

Examine the nature of the roots of the following quadratic equation.

x² - 16x + 64 =0

Solution :

If x² - 16x + 64 =0 is compared to the general form ax² + bx + c =0,

we get a = 1, b = -16 and c = 64.

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = (-16)² - 4(1)(64)

b² - 4ac = 256 - 256

b² - 4ac = 0

Hence, the roots are real, equal and rational.

Let us look at the next example on nature of the roots of a quadratic equation.

Example 4 :

Examine the nature of the roots of the following quadratic equation.

3x² + 5x + 8 =0

Solution :

If 3x² + 5x + 8 =0 is compared to the general form ax² + bx + c =0,

we get a = 3, b = 5 and c = 8.

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = 5² - 4(3)(8)

b² - 4ac = 25- 96

b² - 4ac = -71 (negative)

Hence, the roots are imaginary.

Let us look at the next example on nature of the roots of a quadratic equation.

Example 5 :

If the roots of the equation 2x² + 8x - m³ = 0 are equal , then find the value of "m"

Solution :

If 2x² + 8x - m³ =0 is compared to the general form ax² + bx + c =0,

we get a = 2, b = 8 and c = -m³.

Since the roots are equal, we have

b² - 4ac = 0

8² - 4(2)(-m³) = 0

64 + 8m³ = 0

8m³ = -64

m³ = -8

m³ = (-2)³

m =  - 2

Hence, the value of "m" is "-2".

Let us look at the next example on nature of the roots of a quadratic equation.

Example 6 :

If the roots of the equation x² - (p+4)x + 2p + 5 = 0 are equal , then find the value of "p"

Solution :

If x² - (p+4)x + 2p + 5 = 0 is compared to the general form                  ax² + bx + c =0,

we get a = 1, b = - (p+4) and c = 2p+5

Since the roots are equal, we have

b² - 4ac = 0

[-(p+4)]² - 4(1)(2p+5) = 0

(p+4)² - 8p - 20  = 0

p² + 8p + 16 -8p -20 = 0

p² - 4 =0

p² = 4

p = ± 2

Hence, the value of "p" is " ±2 "

Let us look at the next example on nature of the roots of a quadratic equation.

Example 7 :

If the roots of the equation x² + (2p-1)x + p² = 0 are real , then find the value of "p"

Solution :

If x² + (2p-1)x + p² = 0 is compared to the general form ax² + bx + c =0,

we get a = 1, b = 2p-1 and c = p²

Since the roots are real, we have

b² - 4ac ≥ 0

(2p-1)² - 4(1)(p²) ≥ 0

4p² - 4p +1 -4p² ≥ 0

- 4p +1 ≥ 0

1 ≥ 4p   (or)   4p ≤ 1

p ≤ 1/4

Hence, the value of "p" is less than or equal to "1/4"

Let us look at the next example on nature of the roots of a quadratic equation.

Example 8 :

If the roots of the equation x² - 16x + k =0 are real and equal, then find the value of "k"

Solution :

If x² -16x + k = 0 is compared to the general form ax² + bx + c =0,

we get a = 1, b = -16 and c = k

Since the roots are real, we have

b² - 4ac = 0

(-16)² - 4(1)(k) = 0

64 - 4k = 0

64 = 4k

16 = k

(or) k = 4

Hence, the value of "k" is "4"

Let us look at the next example on nature of the roots of a quadratic equation.

Example 9 :

Examine the nature of the roots of the following quadratic equation.

x² - 5x = 2(3x+1)

Solution :

First, let us write the given equation in general form.

x² - 5x = 2(3x+1)

x² - 5x = 6x+2

x² - 11x -2 = 0

If x² -11x - 2 = 0 is compared to the general form ax² + bx + c =0,

we get a = 1, b = -11 and c = -2

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = (-11)² - 4(1)(-2)

b² - 4ac = 121 + 8

b² - 4ac = 129 (>0, but not a perfect square)

Hence, the roots are real, distinct and irrational.

Let us look at the next example on nature of the roots of a quadratic equation.

Example 10 :

Examine the nature of the roots of the following quadratic equation.

Solution :

If 2x² - 9x -6 = 0 is compared to the general form ax² + bx + c =0,

we get a = 2, b = -9 and c = -6.

Now, let us find the value of the discriminant "b² - 4ac"

b² - 4ac = (-9)² - 4(2)(-6)

b² - 4ac = 81+ 48

b² - 4ac = 129 (>0, but not a perfect square)

Hence, the roots are real, distinct and irrational.

We hope that the student would have understood the problems and solutions given on "nature of the roots of quadratic equations".

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