0 0 how do i find the number of real number solutions to an equation if there is no b value? All the equations I know for finding this need a b value.

Question

U have a huge test retake coming up

The problems I cant find answers for are

-4=x^2-5 and -4=x^2+2x

Answer 1

The first part is to set the equations to 0 by moving everything to one side:

-4 = x² - 5

Here we add 4 to both sides, giving us:

0 = x² - 1

What we have here is a difference of two squares: the square root of x² is x, and the square root of 1 is 1. This means that we can easily factor the problem out with one factor as positive and the other negative:

0 = (x + 1)(x - 1)

So:

0 = x + 1, or 0 = x - 1

x = -1 or 1

When you check your answer putting these values into the original equation, you will see that 1 and -1 both will lead to a final answer of -4.

 

 

-4 = x² + 2x

Again, we start by making the equation equal to 0 by moving everything to one side:

0 = x² + 2x + 4

This equation has no real number solutions. We can take a look at its graph, and we can see the lowest point is going to be at the vertex of (-1, 3). It doesn't cross the x-intercept, so there are no solutions.