Using the Quadratic Formula
The quadratic formula can be used to solve any quadratic equation.
Equation 1
Two rational solutions
Solve x^{2} + 2x  15 = 0 using the quadratic formula.
Solution
To use the formula, we first identify the values of a, b, and c:
The coefficient of x^{2} is 1, so a = 1. The coefficient of 2x is 2, so b
= 2. The constant
term is 15, so c = 15. Substitute these values into the quadratic formula:
Check 3 and 5 in the original equation. The solution set is {5, 3}.
Caution
To identify a, b, and c for the quadratic formula, the equation
must be in the standard form ax^{2} + bx + c = 0. If it is not in that form, then you
must first rewrite the equation.
Example 2
One rational solution
Solve 4x^{2} = 12x  9 by using the quadratic formula.
Solution
Rewrite the equation in the form ax^{2} + bx + c = 0 before identifying a, b, and c:
4x^{2}  12x + 9 = 0
In this form we get a = 4, b = 12, and c = 9.
Check
in the original equation. The solution set is
.
Because the solutions to the equations in Examples 1 and 2 were rational numbers,
these equations could have been solved by factoring. In the next example the
solutions are irrational.
Example 3
Two irrational solutions
Solve 2x^{2} + 6x + 3 = 0.
Solution
Let a = 2, b = 6, and c = 3 in the quadratic formula:
Check these values in the original equation. The solution set is
.
Example 4
Two imaginary solutions, no real solutions
Find the complex solutions to x^{2} + x + 5 = 0.
Solution
Let a = 1, b = 1, and c = 5 in the quadratic formula:
Check these values in the original equation. The solution set is
. There
are no real solutions to the equation.
Number of Solutions to a Quadratic Equation
The quadratic equations in Examples 1 and 3 had two real solutions each. In each of
those examples the value of b^{2}  4ac was positive. In Example 2 the quadratic
equation had only one solution because the value of b^{2}  4ac was zero. In Example
4 the quadratic equation had no real solutions because b^{2}  4ac was negative.
Because b^{2}  4ac determines the kind and number of solutions to a quadratic
equation, it is called the discriminant.
Number of Solutions to a Quadratic Equation
The quadratic equation ax^{2} + bx + c = 0 with a
≠ 0 has
two real solutions if b^{2}  4ac >
0,
one real solution if b^{2}  4ac = 0, and
no real solutions (two imaginary solutions) if b^{2}  4ac < 0.
Example 5
Using the discriminant
Use the discriminant to determine the number of real solutions to each quadratic
equation.
a) x^{2}  3x  5 = 0
b) x^{2} = 3x  9
c) 4x^{2}  12x + 9 = 0
Solution
a) For x^{2}  3x  5 = 0, use a = 1, b = 3, and c = 5 in b^{2}
 4ac:
b^{2}  4ac = (3)^{2}  4(1)(5) = 9 + 20 = 29
Because the discriminant is positive, there are two real solutions to this quadratic
equation.
b) For x^{2}  3x + 9 = 0, use a = 1, b = 3, and c = 9 in b^{2}
 4ac:
b^{2}  4ac = (3)^{2}  4(1)(9) = 9  36 = 27
Because the discriminant is negative, the equation has no real solutions. It has
two imaginary solutions.
c) For 4x^{2}  12x + 9 = 0, use a = 4, b = 12, and c = 9 in b^{2}
 4ac:
b^{2}  4ac = (12)^{2}  4(4)(9) = 144  144 = 0
Because the discriminant is zero, there is only one real solution to this quadratic
equation.
