Solving Using the Quadratic Formula

Each time we solve by completing the square, the procedure is the same. In mathematics, when a procedure is repeated many times, a formula can often be developed to speed up our work.

If we begin with a quadratic equation in standard form, ax2 + bx + c = 0, and solve by completing the square we arrive at the quadratic formula.

Example

Solution

Solve 3x2 + 5x = 2 using the quadratic formula.

First determine a, b, and c:

3x2 + 5x – 2 = 0

a = 3, b = 5, and c = –2.

2 4

2

b b acx

a

25 5 4(3)( 2)

2(3)x

Substituting

1 or 2

3x x

2 12 or

6 6x x

5 7 5 7 or

6 6x x

5 7

6x

5 25 24

6x

5 49

6x

3 21or

3x

Example

Solution

First determine a, b, and c:

x2 – 2x + 7 = 0

a = 1, b = –2, and c = 7.

2 4

2

b b acx

a

22 ( 2) 4(1)(7)

2(1)x

Solve x2 + 7 = 2x using the quadratic formula.

Substituting

2 4 28

2x

2 2 6

2

ix

2 2 6

2 2x i

2 24

2x

1 6x i

Approximating Solutions

When the solution of an equation is irrational, a rational-number approximation is often useful. This is often the case in real-world applications similar to those found in section 8.3.

Example

Solution

2 31.

3

Use a calculator to approximate

2 312.522588.

3

Take the time to familiarize yourself with your calculator: