Algebra Know-It-ALL

692 Worked-Out Solutions to Exercises: Chapters 21 to 29

Because d= 0, we know that the quadratic we get by setting the trinomial equal to

0 has one real root with multiplicity 2. That means the original cubic has one more

real root besides x= 5, and that root has multiplicity 2. To find it, we can use the

quadratic formula with the coefficients and constant named according to the above

scheme:

x= [−b 2 ± (b 22 − 4 a 2 c)1/2] / (2a 2 )

Because the discriminant is equal to 0, we can simplify this to

x=−b 2 / (2a 2 )

Substituting in the values a 2 =−9 and b 2 =−24, we get

x=−(−24) / [2 × (−9)]

= 24/(−18)

=−24/18

=−4/3

The roots of the cubic are therefore x= 5 or x=−4/3. The root x=−4/3 occurs with

multiplicity 2. The solution set is X= {5, −4/3}.

10. The new cubic, written in binomial-trinomial form, looks like this:

(x+ 3/2)(6x^2 − 4 x+ 2) = 0

Let’s examine the discriminant d of the trinomial. Setting a 2 = 6, b 2 =−4, and c= 2,

we get

d=b 22 − 4 a 2 c

= (−4)^2 − 4 × 6 × 2

= 16 − 48

=− 32

Because d < 0, the new cubic has only one real root, x=−3/2, exactly as the original cubic

did. (The two complex roots, however, differ in this equation compared with those in the

final “challenge” in the chapter text. For extra credit, you can verify this fact.)

Chapter 26

1. In each of these situations, the trinomial can be factored into the square of a binomial.
   Then that squared binomial is raised to the indicated power.
   (a) Here is the original equation:

(x^2 + 6 x+ 9)^2 = 0