Question

how many real solutions does the equation have?
$m^2 = 71$
no real solution
one real solution
two real solutions

Explanation:

Step1: Recall the square root property

For an equation of the form \(x^{2}=a\) (where \(a\) is a real number), if \(a > 0\), then \(x=\sqrt{a}\) or \(x =-\sqrt{a}\), so there are two real solutions. If \(a = 0\), then \(x = 0\) (one real solution). If \(a<0\), there are no real solutions.

Step2: Analyze the given equation \(m^{2}=71\)

Here, \(a = 71\), and \(71>0\). So, by the square root property, \(m=\sqrt{71}\) or \(m=-\sqrt{71}\). These are two distinct real numbers.

Answer:

two real solutions