1001 Algebra Problems.PDF

555. c.First, we must make certain that the numer-
     ator and denominator are both completely fac-
     tored and that all common terms are canceled:

^6 x^324 – x^224 x=  46 x( 6 x^2 –x^4 )x= =

Now, the strategy is to determine the x-values

that make this expression equal to zero or

undefined. Then, we assess the sign of the

expression on the left side on each subinterval

formed using these values, which are x= –2, 0,

2. Now, we form a number line, choose a real
   number in each subinterval, and record the
   sign of the expression above each:

Since the inequality includes “equals,” we

include those values from the number line that

make the numerator equal to zero. The solu-

tion set is [–2, 0)∪(2,∞).

556. c.First, we must make certain that the numer-
     ator and denominator are both completely fac-
     tored and that all common terms are canceled:

=

= ^39 = ^13 

Now, determine the x-values that make this

expression equal to zero or undefined. Then,

we assess the sign of the expression on the left

side on each subinterval formed using these

values. To this end, observe that the only value

for which this is true is x= ^52 . Next, we form a

number line, choose a real number in each of

the duly formed subintervals, and record the

sign of the expression above each:

Since the inequality does not include

“equals,”so we do not include those values

from the number line that make the numera-

tor equal to zero. There are no such values, and

furthermore, the expression is always positive.

Therefore, the solution set is the empty set.

557. c.First, make certain that the numerator and
     denominator are both completely factored and
     that all common terms are canceled, as
     follows:

(x+^32 – )(^2 xx–1)– (x–^2 1)–(xx+2)= ^3 (x–^2 – x1)–((2x+– 2 x))=

(x–1^1 )–(xx+2)= (x––( 1 x)(–x^1 +)2)= –x+^1  2

Next, determine the x-values that make this

expression equal to zero or undefined. Then,

we assess the sign of the expression on the left

side on each subinterval formed using these

values, which are x= –2 and 1. Now, we form a

number line, choose a real number in each

subinterval, and record the sign of the expres-

sion above each:

Since the inequality includes “equals,” so we

include those values from the number line that

make the numerator equal to zero. The solu-

tion set is (–∞, –2).

–2 1

+––

5

2

(2x– 5)((x+ 4) – (x+ 1))

9(2x– 5)

(2x– 5)(x+ 4) – (2x– 5)(x+ 1)

9(2x– 5)

• 2 0 2


• 
  
  
  
  • – +
  
  
  
  

(x– 2)(x+ 2)

4 x

x^2 – 4

4 x

ANSWERS & EXPLANATIONS–