Question

The equation x^2-9=0 has how many real solutions

Answer 1

So remember that the degree of the polynomial will indicate how many solutions there are, real or complex. Since the degree of the polynomial is 2, this means that there is a maximum of 2 real solutions in this equation.

Next, we are going to apply the difference of squares rule to this polynomial, which is $x^2-y^2=(x+y)(x-y)$ . In this case:

$x^2-9=(x+3)(x-3)\\(x+3)(x-3)=0$

Now, we are going to apply the Zero Product Property, which states that if a × b = 0, this means that either a or b = 0 or a and b = 0. In this case, a = x + 3 and b = x - 3. Solve each as such:

$x+3=0\\x=-3\\\\x-3=0\\x=3$

In short, there are 2 real solutions: x = 3 and x = -3.