1001 Algebra Problems.PDF

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

^2 z 2 z^2 +– 2 z z––^1515 = ((^2 zz++ 55 ))((zz–– 33 ))= (2zz++ 5 5)

Now, the strategy is to determine the z-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 these values arez= –5, –^52 . Next, 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,” we

include those values from the number line that

make the numerator equal to zero. Therefore,

the solution set is (–∞, –5)∪[–^52 ,∞).

550. d.Our first step is to make certain that the
     numerator and denominator are both com-
     pletely factored and that all common terms are
     canceled:

= = ^1 x

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= 0. Next, we form a

number line, choose a real number in each

subinterval, and record the sign of the expres-

sion above each, as follows:

Since, the inequality includes “equals,” so include

those values from the number line that make

the numerator equal to zero. Since none of

these values make the numerator equal to zero,

we conclude that the solution set is (–∞, 0).

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

z 83 z––^1362 z= z( 8 z(^2 z––^146 ))= z(z 8 – (^4 z)–(z 4 +)4)=

Next, determine the z-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 these values

arez= –4, 0, 4. Now, 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,”

we do not include those values from the num-

ber line that make the numerator equal to

zero. The solution set is (–4, 0).

4 0 4

+ – + +

z(z+4)

8

0

^25 x^4

x 25 x^4

25(–x)^4

x(5x^2 )^2

(^552)

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ANSWERS & EXPLANATIONS–