Question

what are the possible values of a such that $a^2b = 12$, $a < 20$, $b < 20$, and a and b are integers?
a. -1, 1
b. -2, 2
c. -1, 1, -2, 2
d. -1, 1, -3, 3

when we plug in 2 for a, we get
$2^2b = 12$
$4b = 12$

is 2 a possible value of a?
a. yes, since when a is 2, b is an integer less than 20.
b. no, since when a is 2, b is not an integer less than 20.

Explanation:

First Question (Multiple Choice)

Step1: Rearrange for b

$b = \frac{12}{a^2}$

Step2: Test integer a values

For $a=\pm1$: $b=\frac{12}{1^2}=12$ (integer, $12<20$)
For $a=\pm2$: $b=\frac{12}{2^2}=3$ (integer, $3<20$)
For $a=\pm3$: $b=\frac{12}{3^2}=\frac{4}{3}$ (not integer)

Step1: Solve for b when a=2

$2^2b=12 \implies 4b=12 \implies b=3$

Step2: Check conditions

$b=3$ is integer, $3<20$

Answer:

C. -1, 1, -2, 2

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Second Question (Yes/No)