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

Step-by-step 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.