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. $-1, 1, -3, 3$
which equation do you get by replacing a with 2?
a $2b = 12$
b $4b = 12$
c $2a^2 = 12$

Explanation:

First Question

Step1: List integer squares dividing 12

The divisors of 12 that are perfect squares are $1, 4$ (since $9$ does not divide 12, $16>12$).

Step2: Find $a$ for each square

For $a^2=1$: $a=\pm1$. Calculate $b$: $b=\frac{12}{1}=12$, which is $<20$.
For $a^2=4$: $a=\pm2$. Calculate $b$: $b=\frac{12}{4}=3$, which is $<20$.

Step3: Verify invalid $a$ values

If $a=\pm3$, $a^2=9$, $b=\frac{12}{9}=\frac{4}{3}$, not an integer, so invalid.

Step1: Substitute $a=2$ into $a^2b=12$

Calculate $a^2$ when $a=2$: $2^2=4$. Substitute into the equation.

Answer:

C. -1, 1, -2, 2

---

Second Question