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Using the Quadratic Formula

The quadratic formula can be used to solve any quadratic equation.

 

Equation 1

Two rational solutions

Solve x2 + 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 x2 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 ax2 + bx + c = 0. If it is not in that form, then you must first rewrite the equation.

 

Example 2

One rational solution

Solve 4x2 = 12x - 9 by using the quadratic formula.

Solution

Rewrite the equation in the form ax2 + bx + c = 0 before identifying a, b, and c: 4x2 - 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 2x2 + 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 x2 + 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 b2 - 4ac was positive. In Example 2 the quadratic equation had only one solution because the value of b2 - 4ac was zero. In Example 4 the quadratic equation had no real solutions because b2 - 4ac was negative. Because b2 - 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 ax2 + bx + c = 0 with a 0 has

two real solutions if b2 - 4ac > 0,

one real solution if b2 - 4ac = 0, and

no real solutions (two imaginary solutions) if b2 - 4ac < 0.

 

Example 5

Using the discriminant

Use the discriminant to determine the number of real solutions to each quadratic equation.

a) x2 - 3x - 5 = 0

b) x2 = 3x - 9

c) 4x2 - 12x + 9 = 0

Solution

a) For x2 - 3x - 5 = 0, use a = 1, b = -3, and c = -5 in b2 - 4ac:

b2 - 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 x2 - 3x + 9 = 0, use a = 1, b = -3, and c = 9 in b2 - 4ac:

b2 - 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 4x2 - 12x + 9 = 0, use a = 4, b = -12, and c = 9 in b2 - 4ac:

b2 - 4ac = (-12)2 - 4(4)(9) = 144 - 144 = 0

Because the discriminant is zero, there is only one real solution to this quadratic equation.

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