1999 CEMC Pascal Problems/Problem 3

Problem

If $4(\heartsuit)^2 = 144$ , then a value of $\heartsuit$ is

$\text{ (A) }\ 3 \qquad\text{ (B) }\ 6 \qquad\text{ (C) }\ 9 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 18$

We can just treat the heart as a variable like $x$ :

$4x^2 = 144$

Dividing both sides by $4$ , we get: $\frac{4x^2}{4} = x^2 = \frac{144}{4} = 36$

For this to be true, $x = 6$ or $x = -6$ because $6^2 = 6 \times 6 = 36$ , and $(-6)^2 = -6 \times -6 = 36$ .

Of these values for $x$ , $\boxed {\textbf {(B) } 6}$ is an answer choice.

~anabel.disher