376 Quadratic Equations with Real Roots
Because a≠ 0 in any quadratic equation, we can divide each side by a, getting
x^2 + (b/a)x=−c/a
It’s tempting to think that there must be some constant that we can add to both sides of this
equation to get a perfect square on the left side of the equals sign. It takes some searching, but
that constant does exist. It is b^2 /(4a^2 ). When we add it to both sides of the above equation,
we obtain
x^2 + (b/a)x+b^2 /(4a^2 )=−c/a+b^2 /(4a^2 )
We can now factor the left side into the square of a binomial to get
[x+b/(2a)]^2 =−c/a+b^2 /(4a^2 )
The two terms in the right side of this equation can be added using the sum of quotients rule
from Chap. 9 to obtain
[x+b/(2a)]^2 = (− 4 a^2 c+ab^2 ) / (4a^3 )
Canceling out the extra factors of a in the numerator and denominator on the right side, we get
[x+b/(2a)]^2 = (− 4 ac+b^2 ) / (4a^2 )
Let’s rewrite the numerator on the right side as a difference, so the equation becomes
[x+b/(2a)]^2 = (b^2 − 4 ac) / (4a^2 )
If we take the square root of both sides here, remembering the negative as well as the positive,
we get
x+b/(2a)=±[(b^2 − 4 ac) / (4a^2 )]1/2
The denominator in the right side is a perfect square; it’s equal to (2a)^2. Therefore, we can
simplify the expression on that side of the equals sign a little bit, considering it as a ratio of
square roots rather than the square root of a ratio. We obtain
x+b/(2a)=±(b^2 − 4 ac)1/2 / (2a)
If we subtract b/(2a) from both sides, we get
x=±(b^2 − 4 ac)1/2 / (2a)−b/(2a)
which expresses x in terms of the constants a,b, and c (finally!). An equation that states the
general solution to an unknown is called a formula.