Question

The equation of a circle is pi(2x+3)^2

What is the least possible integer value of x for the circle to exist? Please explain.

Answer 1

Answer: x = -1

Explanation:

The radius needs to be positive, so 2x + 3 > 0. Solving for x, we get x > -3/2. The smallest integer larger than -3/2 is -1, so x = –1.