# Free Mathematics Tutorials # Solve Quadratic Equations Using Discriminants

Several questions on how to solve quadratic equations using the discriminant and the quadratic formula are presented along with detailed solutions. We also discuss the relationship between the number and nature of solutions of a given quadratic equation and the sign of its discriminant. More questions with answers are at the bottom of this page.
Also included in this website, a Step by Step Quadratic Equation Solver.

## 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 complex solutions. (see example 3 below)

## Questions with Solutions

Question 1
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x 2 + 3 x = 4
Solution to Question 1

• 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.

Question 2
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x 2 / 3 + 3 = 2 x
Solution to Question 2

• 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 solution given by
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.

Question 3
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x 2 - 4 x + 13 = 0 Solution to Question 3

• 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.

Question 4
Find all values of the parameter m in the quadratic equation
x 2 + m x + 1 = 0
such that the equation has

1. one solution,
2. 2 real solutions,
3. 2 complex solutions.
Solution to Question 4
• a) Given
x 2 + m x + 1 = 0
• Find the discriminant Δ = b2 - 4ac
Δ = b2 - 4ac = m2 - 4(1)(1) = m2 - 4
• For the equation to have one solution, the discriminant has to be equal to zero.
m2 - 4 = 0
• The equation m2 - 4 = 0 has two solutions.
m = 2
m = -2
Below is the graph of the expression in the left side of the given equation for m = 2 and m = -2.
Note that in each case, the graph has 1 x intercept only, hence one real solution to the equation. • b) For the equation to have 2 real solution, the discriminant has to be greater than zero.
m2 - 4 > 0
• The inequality m2 - 4 > 0 has the following solution set.
(-∞ , -2) U (2 , +∞) &infty;
Below is the graph of the expression in the left side of the given equation for m = 5 and m = -3.
Note that in each case, the graph has 2 x-intercepts, hence the given equation has 2 real solutions. • c) For the equation to have 2 complex solution, the discriminant has to be less than zero.
m2 - 4 < 0
• The inequality m2 - 4 < 0 has the following solution set.
(-2 , 2)

Below is the graph of the expression in the left side of the given equation for m = 0 and m = 1.
Note that in each case, the graph has no x-intercepts, hence the solutions to the equation are not real but complex. ## More Questions with Detailed Solutions

1) Use the quadratic formulas and the discriminant to find all solutions to the quadratic equations given below.
a) x 2 - 3 x + 2 = 0

b)
x 2 / 2 = - 8 - 4x

Detailed Solution

c)
x 2 - 4 x + 5 = 0

Detailed Solution

2) Find all values of the parameter m in the quadratic equation
x 2 + x + m + 1 = 0
for each case below:
1. the equations has one solution,
2. the equations has 2 real solutions,
3. the equations has 2 complex solutions.

## More Questions (with answers below)

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