# Solve Quadratic Equations Using Discriminants (1)

This is a tutorial on using the discriminant and the quadratic formula to solve quadratic equations. We also discuss the relationship between the number of solutions the value of the discriminant. Detailed solutions and explanations to questions are included. More questions with answers are at the bottom of this page.

## Review of the Quadratic Formulas and Discriminant

A quadratic equation in one variable is an equation that may be written in the form

a x 2 + b x + c = 0

where a, b and c are constants with a not equal to zero.

There are several methods to solve quadratic equations. In this tutorial we use the the quadratic formulas and the discriminant.
The solutions to the above equation are given by the quadratic formulas.

x1 = [ - b + √(b2 - 4 a c) ] / (2 a)
and
x2 = [ - b - √(b2 - 4 a c) ] / (2 a)

The quantity Δ = b2 - 4 a c under the radical above is called the discriminant and gives important information about the number and nature of the solutions to the quadratic equation to be solved. Three cases are possible:
1. If Δ > 0, the equation has 2 real solutions. (see example 1 below)
2. If Δ = 0, the equation has 1 real solution. (see example 2 below)
3. If Δ < 0, the equation has 2 conjugate imaginary solutions. (see example 3 below)

### Question1:

Find all solutions to the quadratic equation given below.

x 2 + 3 x = 4

### Solution to Question1:

• Given
x 2 + 3 x = 4
• Rewrite the given equation with its right term equal to zero.
x 2 + 3 x - 4 = 0
• Find the discriminant Δ= b2 - 4 a c
Δ= b2 - 4 a c = 32 - 4(1)(-4) = 25
• Since the discriminant is positive, the equation has two real solutions given by.
x1 = [-3 + √25] / (2*1) = [-3 + 5] / 2 = 1
x2 = [-3 - √25] / (2*1) = [-3 - 5] / 2 = - 4

### Check Solutions

1. x = 1
Left side of the equation = x 2 + 3 x = 1 2 + 3(1) = 1 + 3 = 4
Right side of the equation = 4.
2. x = -4
Left side of the equation = (-4) 2 + 3(-4) = 16 - 12 = 4
Right side of the equation = 4.

Conclusion: The solutions to the given equation are 1 and -4.

### Matched Question 1:

Find all solutions to the quadratic equation given below.

x 2 - 3 x + 2 = 0

### Question2 :

Find all solutions to the quadratic equation

x 2 / 3 + 3 = 2 x

### Solution to Question2:

• Given
x 2 / 3 + 3 = 2 x
• Eliminate the denominator by multiplying all terms in the equation by 3.
3[x 2 / 3 + 3] = 3*2x
• Simplify and rewrite the equation with the right term equal to zero.
x 2 - 6 x + 9 = 0
• Use the quadratic formula. The discriminant Δis given by
Δ= b2 - 4 a c
= (-6)2 - 4(1)(9) = 0
• Since the discriminant is equal to zero, the two formulas giving the two solutions of the quadratic equation become one x = -b/2a and the equation has one solution.
x = -b / (2 a) = -(-6) / (2*1) = 3

### Check Solutions

1. x = 3
Left side of the equation = x 2 / 3 + 3 = 3 2 / 3 + 3 = 6
Right side of the equation = 2 x = 2(3) = 6.

Conclusion:
There is one real solution to the given equation: x = 3.

Matched Question 2. Find all solutions to the quadratic equation.

x 2 / 2 = - 8 - 4x

### Question3:

Find all solutions to the quadratic equation

x 2 - 4 x + 13 = 0

### Solution to Question3:

• Given
x 2 - 4 x + 13 = 0
• The discriminant Δis given by
Δ= b2 - 4 a c
= (-4)2 - 4(1)(13) = -36

• Since the discriminant is negative, the square root of the discriminant is a pure imaginary number.
sqrt(D) = √(-36) = √(-1)√(36) = 6i
where i is the imaginary unit defined as i = √(-1).
• Use the quadratic formulas to find the two solutions.
x1 = (4 + 6 i)) / (2*1) = 2 + 3 i
x2 = (4 - 6 i) / (2*1) = 2 - 3 i

### Check Solutions

1. x = 2 + 3 i
Left side of the equation = x 2 - 4 x + 13 = (2 + 3 i) 2 - 4(2 + 3 i) + 13
= 4 - 9 + 12 i - 8 - 12 i + 13 = 0
Right side of the equation = 0
2. x = 2 - 3 i
Left side of the equation = x 2 - 4 x + 13 = (2 - 3 i) 2 - 4(2 - 3 i) + 13
= 4 - 9 - 12 i - 8 + 12 i + 13 = 0
Right side of the equation = 0

Conclusion:
The given equation has two imaginary solutions 2 + 3 i and 2 - 3 i conjugate of each other.

### Matched Question 3.

Find all solutions to the quadratic equation.

x 2 - 4 x + 5 = 0

### More Questions.

a) - x
2 + 2 x = - 3
b) (1 / 2) x
2 + (1 / 3) x = 1 / 6
c) x
2 + 9 = 0
d) - 0.2 x
2 + 2.0 x = + 5.2
e) [ 3 x
2 + 2x ] / 2 = 2

a) -1 , 3
b) -1 , 1 / 3
c) 3 i , -3 i
d) 5 - i , 5 + i
e) √(13) / 3 - 1 / 3 , - √(13) / 3 - 1 / 3