5.4 The Quadratic Formula - College of the Redwoods

484 Chapter 5 Quadratic Functions
Next, divide both sides of the equation by a.
x 2 + b a x = − c a
Take half of the coefficient of x, as in (1/2)(b/a) = b/(2a). Square this result to get
b 2 /(4a 2 ). Add this amount to both sides of the equation.
x 2 + b a x + b2
4a 2 = − c a + b2
4a 2
On the left we factor the perfect square trinomial. On the right we get a common
denominator and add the resulting equivalent fractions.
(
x + b ) 2
= − 4ac
2a 4a 2 + b2
4a 2
(
x + b ) 2
= b2 − 4ac
2a 4a 2
Provided the right-hand side of this last equation is positive, we have two real solutions.
x + b
√
b
2a = ± 2 − 4ac
4a 2
On the right, we take the square root of the top and the bottom of the fraction. 6
x + b
√
b
2a = ± 2 − 4ac
2a
To complete the solution, we need only subtract b/(2a) from both sides of the equation.
x = − b
√
b
2a ± 2 − 4ac
2a
Although this last answer is a perfectly good solution, we customarily rewrite the
solution with a single common denominator.
x = −b ± √ b 2 − 4ac
2a
This last result gives the solution to the general quadratic equation (8). The solution
(9) is called the quadratic formula.
(9)
6 In a later section we will present a more formal approach to the symbolic manipulation of radicals.
For now, you can compute (2/3) 2 with the calculation (2/3)(2/3) = 4/9, or you can simply square
numerator and denominator of the fraction, as in (2/3) 2 = (2 2 /3 2 ) = 4/9. Conversely, one can take the
square root of a fraction by taking the square root of the numerator divided by the square root of the
denominator, as in √ 4/9 = √ 4/ √ 9 = 2/3.
Version: Fall 2007