# Algebra II

posted by Anonymous

How many real solutions does the function shown on the graph have?

a) no real solutions
b) one real solution
c) two real solutions
d) cannot be determined
___________________________
y = (x+2)^2 + 3 is shown on the graph.

I am having trouble figuring this one out. I have been looking in my book for a while now. Somebody please help, it would be very much appreciated!

1. Anonymous

I think it could be a) no real solutions?
Am I correct? If not, please explain...

2. Anonymous

Whoops. I meant the graph was showing: y=(x+2)^2+2

3. Anonymous

I was thinking that it is a) because it does not have any x-intercepts?

4. Bosnian

( x + 2 )² + 2 = 0

Subtract 2 to both sides

( x + 2 )² + 2 - 2 = 0 - 2

( x + 2 )²= - 2

Take square root of both sides

x + 2 = ± √( - 2 )

x + 2 = ± √( - 1 ∙ 2 )

x + 2 = ±√( - 1 ) ∙ √2

x + 2 = ± i √2

Subtract 2 to both sides

x + 2 - 2 = ± i √2 - 2

x = - 2 ± i √2

The solutions are:

x = - 2 + i √2 and x = - 2 - i √2

The function ( x + 2 )² + 2 no real solutions.

5. Reiny

Just looking at the equation, we can tell that the vertex is (-2,2) and the parabola opens upwards. So clearly, it cannot cross the x-axis. So clearly, no real solution

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