Asked 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!
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!
Answers
Answered by
Anonymous
I think it could be a) no real solutions?
Am I correct? If not, please explain...
Am I correct? If not, please explain...
Answered by
Anonymous
Whoops. I meant the graph was showing: y=(x+2)^2+2
Answered by
Anonymous
I was thinking that it is a) because it does not have any x-intercepts?
Answered by
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.
Answer a) is correct
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
Answered by
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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