Algebra Know-It-ALL

520 Review Questions and Answers

Answer 26-3

Annth-degree equation in binomial-to-the-nth form can always be written like this when x

is the variable:

(ax+b)n= 0

where a is a nonzero real number, b is a real number, and n is a positive integer. Theoretically,

n can be any positive integer. If n= 1, the equation is of the first degree; if n= 2, the equation

is quadratic; if n= 3, the equation is cubic. If n > 3, the equation is of higher degree.

Question 26-4

How many real roots does the equation stated in Answer 26-3 have? What is that root, or what

are they? What is the real solution set X?

Answer 26-4

There is one real root with multiplicity n. It is the solution to the equation we obtain when

we set the binomial equal to 0:

ax+b= 0

When we subtract b from both sides and then divide through by a (which is okay because we

know that a≠ 0), we get

x=−b/a

The real solution set is therefore

X= {−b/a}

Question 26-5

Find all the real roots of the following equation, state the multiplicity of each, and state the

real solution set X.

(x^2 − 6 x+ 9)^2 = 0

Answer 26-5

Let’s set the quantity (x^2 − 6 x+ 9) equal to 0, so it becomes the quadratic equation

x^2 − 6 x+ 9 = 0

We can factor this into

(x− 3)^2 = 0