# NATURE OF THE ROOTS OF A QUADRATIC EQUATION EXAMPLES

Nature of the roots of a quadratic equation examples are nothing but some sample problems on the above stuff.

Before do the problems, we have to know the following stuff.

Based on the stuff given above, we can solve any problem on nature of the roots of a quadratic equation examples.

## Examples

Now, let us do some problems on "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 problem on "Nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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.

Now we are going to look at some quiet different problems on "nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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 -4 ≥ 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 problem on "Nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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 problem on "Nature of the roots of a quadratic equation examples"

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 a quadratic equation examples".

Apart from nature of the roots of a quadratic equation examples, you can also visit the following web pages to know more about quadratic equation.

Related Topics  :

1. Nature of the roots of a quadratic equation (Detailed stuff)

2. Relationship between zeros and coefficients of a quadratic polynomial

3. Word problems on quadratic equation

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