Algebra Know-It-ALL

Part Three 503

Answer 22-4

This equation is a product of binomials. The roots are the values of x that make either factor
equal to 0. We can find those roots by setting each binomial equal to 0, creating two separate
first-degree equations. Then we can morph the equations to get the variable all alone on the
left side of the equals sign, and some combination of the coefficient and constant on the right
side. For the first term, the process goes like this:

a 1 x+b 1 = 0

a 1 x=−b 1

x=−b 1 /a 1

For the second term, it’s the same, but with subscripts of 2 instead of 1:

a 2 x+b 2 = 0

a 2 x=−b 2

x=−b 2 /a 2

Question 22-5

How can we solve the following quadratic equation, where a and b are real numbers, and x is
the variable? Assume that a≠ 0:

(ax+b)^2 = 0

Answer 22-5

We can look at this equation as a product of two identical binomials:

(ax+b)(ax+b)= 0

We can solve this quadratic in the same way as we solved the equation in Answer 22-4. But
this time, we get only one root:

ax+b= 0

ax=−b

x=−b/a

Question 22-6

What’s special about the root of the equation we solved in Answer 22-5?

Answer 22-6

This root, −b/a, occurs “twice over.” In technical terms, the root has multiplicity 2.

Question 22-7

Suppose we see a quadratic equation in polynomial standard form. Again, here’s the general form:

ax^2 +bx+c= 0