Algebra Know-It-ALL

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.