nature of the solutions

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Nature of the Solutions Value of the discriminant Type and number of Solutions Example of graph Positive Discriminant b² − 4ac > 0 Two Real Solutions If the discriminant is a perect square the roots are rational. Otherwise, they are irrational. Discriminant is Zero b² − 4ac = 0 One Real Solution Negative Discriminant b² − 4ac < 0 No Real Solutions Two Imaginary Solutions Example 1 Quadratic Equation: y = x² + 2x + 1 o a = 1 o b = 2 o c = 1 The discriminant for this equation is 2² - 4•1 •1= 4 − 4 = 0 Since the discriminant is zero , there should be 1 real solution to this equation. Below is a picture representing the graph and one solution of this quadratic equation Graph of y = x² + 2x + 1

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Nature of the Solutions

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Nature of the Solutions

Nature of the Solutions

Value of the discriminant Type and number of Solutions Example of graph

Positive Discriminant

b 4ac > 0 Two Real Solutions

If the discriminant is a perect square the roots are rational. Otherwise, they are irrational.

Discriminant is Zero

b 4ac = 0 One Real Solution

Negative Discriminant

b 4ac < 0 No Real Solutions Two Imaginary Solutions

Example 1 Quadratic Equation: y = x + 2x + 1 a = 1

b = 2

c = 1

The discriminant for this equation is

2 - 41 1= 4 4 = 0

Since the discriminant is zero, there should be 1 real solution to this equation. Below is a picture representing the graph and one solution of this quadratic equation

Graph of y = x + 2x + 1

Calculate the discriminant to determine the number and nature of the solutions of the following quadratic equation:

y = x 2x + 1

Answer

In this quadratic equation,

y = x 2x + 1

a =1

b = 2

c = 1

Using our general formula, the discriminant is (-2) 41 1 = 4 4 = 0 Since the discriminant is zero, we should expect 1 real solution which you can see pictured in the graph below.

Use the discriminant to find out the nature and number of solutions:

y = x x 2

Answer

In this quadratic equation,

y = x x 2 and its solution

a =1

b = 1

c = 2

Discriminant: b 4(a)(c) = (-1) 4(1)(-2) 1 -8 = 1 + 8 = 9

Since the discriminant is positive, there should be 2 real solutions to this equation.

Calculate the discriminant to determine the nature and number of solutions:

y = x 1

Answer

In this quadratic equation,

y = x 1

a = 1

b = 0

c = 1

Discriminant: b 4(a)(c) = (0) 4(1)(-1) -4 = 4

Since the discriminant is positive, we have two real solutions. The exact solutions are show below.

Calculate the discriminant to determine the nature and number of solutions:

y = x + 4x + 5

Answer

In this quadratic equation,

y = x + 4x + 5

a =1

b = 4

c = 5

the discriminant = b 4(a)(c) = 4 4(1)(5) 16 20 = - 4

Since the discriminant is negative , there are noreal solutions to this quadratic equation. The only solutions are imaginary.

Below is a picture of this quadratic's graph

Find the discriminant to determine the nature and number of solutions:

y = x + 4

Answer

y = x + 4

a =1

b = 0

c = 4

the discriminant = b 4(a)(c) = 0 4(1)(4) 16

Since the discriminant is negative , there are two imaginary solutions to this quadratic equation.

The solutions are 2i and -2i Below is a a picture of this equations graph Find the discriminant to determine the nature and number of solutions:

y = x + 25

Answer

y = x + 25

a =1

b = 0

c = 25

the discriminant = b 4(a)(c) = 0 4(1)(25) 100 = 100

Since the discriminant is negative , there are two imaginary solutions to this quadratic equation.

The solutions are 5i and -5i