Question

how many real solutions does the equation have?
$s^2 + 32 = 0$
no real solution
one real solution
two real solutions

Explanation:

Step1: Analyze the equation \(s^2 + 32 = 0\)

We can rewrite the equation as \(s^2=- 32\).

Step2: Recall the property of square of real numbers

For any real number \(s\), the square of \(s\) (i.e., \(s^2\)) is always non - negative. That is, if \(s\in\mathbb{R}\), then \(s^2\geq0\). But in our equation, we have \(s^2=-32\), and \(- 32<0\). So there is no real number \(s\) that satisfies the equation \(s^2 + 32=0\).

Answer:

no real solution