Question

How many real solutions does the equation have?
g² +42 = 0

Answer 1

First, isolate “g” by subtracting 42 from both sides:

g²=0-42

g²=-42

Now, we can take the square root of both sides to cancel out the g²:

±√g²=±√-42

We have a problem here: negative roots do not exist. This is because no two same-signed values multiply together to get -42. View this example: a²=a•a and a²=-a•-a. A positive multiplied to a positive value produces a positive value, and a negative multiplied by a negative produces a positive value. By the definition of exponents, a number multiplied to itself can only be the same-sign as its base. Therefore, we have two imaginary solutions. The rule for complex numbers is: i=√-1 and i²=-1.

So, let’s simplify the equation into the proper complex numbers:

g=±√42•-1

Notice how the 42 is factored into 42•-1. This is because √-1=i.

g=±i√42

Thus, the equation has no real solutions.

Answer 2

The equation g² + 42 = 0 has two real solutions.

To solve this equation, we can start by subtracting 42 from both sides to get g² = -42. Then, we can take the square root of both sides to get |g| = √42. Since the square root of a number is positive, this means that g = √42 or g = -√42. These are the two solutions to the equation.