# 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 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 + s=  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. Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

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