NATURE OF THE ROOTS OF A QUADRATIC EQUATION WORKSHEET

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

Detailed Answer Key

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.  

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