NATURE OF THE ROOTS OF A QUADRATIC EQUATION WORKSHEET

Nature of the Roots of a Quadratic Equation Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice problems on nature of the roots of a quadratic equation.

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Nature of the Roots of Quadratic Equation Worksheet - Problems

Problem 1 :

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

x² + 5x + 6 = 0

Problem 2 :

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

2x² - 3x - 1 = 0

Problem 3 :

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

x² - 16x + 64 = 0

Problem 4 :

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

3x² + 5x + 8 = 0

Problem 5 :

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

Problem 6 :

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

Problem 7 :

If the roots of the equation x² + (2s - 1)x + s²= 0 are real, then find the value of a.

Problem 8 :

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

Problem 9 :

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

x² - 5x = 2(5x + 1)

Problem 10 :

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

1/(x+1) + 2/(x-4) = 2

Nature of the Roots of Quadratic Equation Worksheet - Solutions

Problem 1 :

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

x² + 5x + 6 = 0

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

Then, we have a = 1, b = 5 and c = 6.

Find the value of the discriminant b² - 4ac.

b² - 4ac = 5² - 4(1)(6)

b² - 4ac = 25 - 24

b² - 4ac = 1

Here, b² - 4ac > 0 and also a perfect square.

So, the roots are real, unequal and rational.

Problem 2 :

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

2x² - 3x - 1 = 0

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

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

Find the value of the discriminant b² - 4ac.

b² - 4ac = (-3)² - 4(2)(-1)

b² - 4ac = 9 + 8

b² - 4ac = 17

Here, b² - 4ac > 0, but not a perfect square.

So, the roots are real, unequal and irrational.

Problem 3 :

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

x² - 16x + 64 = 0

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

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

Find the value of the discriminant b² - 4ac.

b² - 4ac = (-16)² - 4(1)(64)

b² - 4ac = 256 - 256

b² - 4ac = 0

So, the roots are real, equal and rational.

Problem 4 :

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

3x² + 5x + 8 = 0

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

Then, we have a = 3, b = 5 and c = 8.

Find the value of the discriminant b² - 4ac.

b² - 4ac = 5² - 4(3)(8)

b² - 4ac = 25 + 96

b² - 4ac = 121

Here, b² - 4ac > 0 and also a perfect square.

So, the roots are real, unequal and rational.

Problem 5 :

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

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

Then, we have a = 2, b = 8 and c = -m³.

Because the roots of the given equation are equal,

b² - 4ac = 0

8² - 4(2)(-m³) = 0

64 + 8m³ = 0

Subtract 64 from each side.

8m³ = -64

Divide each side by 8.

m³ = -8

m³ = (-2)³

m = -2.

So, the value of m is -2.

Problem 6 :

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

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

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

Because the roots of the given equation are equal,

b² - 4ac = 0

[-(p + 4)]² - 4(1)(2p + 5) = 0

Simplify.

(p + 4)² - 4(2p + 5) = 0

p² + 8p + 16 -8p -20 = 0

p² - 4 = 0

p² - 2² = 0

(p + 2)(p - 2) = 0

p + 2 = 0 or p - 2 = 0

p = -2 or p = 2

So, the value of p is ±2.

Problem 7 :

If the roots of the equation x² + (2s - 1)x + s²= 0 are real, then find the value of a.

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

Then, we have a = 1, b = (2s - 1) and c = s².

Because the roots of the given equation are equal,

b² - 4ac ≥ 0

(2s - 1)² - 4(1)(s²) ≥ 0

Simplify.

4s² - 4s + 1 - 4s² ≥ 0

-4s + 1 ≥ 0

-4s ≥ -1

Divide each side by -4.

s ≤ 1/4

So, the value of s is less than or equal to 1/4.

Note :

Whenever we multiply or divide both sides of an inequality by a negative number, we have to flip the inequality sign.

Problem 8 :

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

Solution :

The given quadratic equation is in the general form

ax² + bx + c = 0

Then, we have a = 1, b = -16 and c = k.

Because the roots of the given equation are equal,

b² - 4ac = 0

(-16)² - 4(1)(k) = 0

256 - 4k = 0

Subtract 256 from each side.

-4k = -256

Divide each side by -4.

k = 64

So, the value of k is 64.

Problem 9 :

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

x² - 5x = 2(5x + 1)

Solution :

The given quadratic equation is not in the general form.

First, write the given quadratic equation in the general form.

x² - 5x = 2(5x + 1)

x² - 5x = 10x + 2

x² - 15x - 2 = 0

Now, the quadratic equation is the general form

ax² + bx + c = 0

Then, we have a = 1, b = -15 and c = -2.

Find the value of the discriminant b² - 4ac.

b² - 4ac = (-15)² - 4(1)(-2)

b² - 4ac = 225 + 8

b² - 4ac = 233

Here, b² - 4ac > 0, but not a perfect square.

So, the roots are real, unequal and irrational.

Problem 10 :

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

1/(x+1) + 2/(x-4) = 2

Solution :

The given quadratic equation is not in the general form.

First, write the given quadratic equation in the general form.

1/(x+1) + 2/(x-4) = 2

Add the two fractions on the right side of the equation using cross multiplication.

[(x-4) + 2(x+1)] / [(x+1)(x-4)] = 2

(x - 4 + 2x + 2) / (x² - 3x - 4) = 2

(3x - 2) / (x² - 3x - 4) = 2

Multiply each side by (x² - 3x - 4).

3x - 2 = 2(x² - 3x - 4)

3x - 2 = 2x² - 6x - 8

2x² - 9x - 6 = 0

Now, the quadratic equation is the general form

ax² + bx + c = 0

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

Find the value of the discriminant b² - 4ac.

b² - 4ac = (-9)² - 4(2)(-6)

b² - 4ac = 81 + 48

b² - 4ac = 129

Here, b² - 4ac > 0, but not a perfect square.

So, the roots are real, unequal and irrational.

After having gone through the stuff given above, we hope that the students would have understood, how to examine the nature of the roots of a quadratic equation.

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