Algebra II

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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!

  • Algebra II -

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

  • Algebra II -

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

  • Algebra II -

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

  • Algebra II -

    ( 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.

    Answer a) is correct

  • Algebra II -

    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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