SOLUTION: Explain why the solution of x^2 + 4x + 8 = 0 does not exist graphically. How do I do this? Is it something about real numbers and complex numbers? Thanks!

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Question 1019490: Explain why the solution of x^2 + 4x + 8 = 0 does not exist
graphically.
How do I do this? Is it something about real numbers and
complex numbers? Thanks!

Found 3 solutions by Edwin McCravy, richard1234, ikleyn:
Answer by Edwin McCravy(20066)   (Show Source): You can put this solution on YOUR website!
Yes, it's about real and complex numbers.



If it had a real solution then they would be 
the x-intercepts of the graph of



For the x-intercepts are when y = 0.  But as you see
when looking at the graph of 

 

It doesn't intercept the x-axis at all!  That means

that the value of y can never be 0.  You can also
tell that it has no real solutions by the quadratic
formula:



a=1, b=4, c=8

















The solutions are complex numbers, and do not 
appear on the graph of 

Edwin

Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
If you graph it, you will see that it lies entirely above the x-axis and does not intersect the x-axis. Therefore x^2 + 4x + 8 has no real solutions (it has two complex solutions).
Answer by ikleyn(52932)   (Show Source): You can put this solution on YOUR website!
.
Explain why the solution of x^2 + 4x + 8 = 0 does not exist
graphically.
How do I do this? Is it something about real numbers and
complex numbers? Thanks!
--------------------------------------------------------

Complete a square for the polynomial

= .

You have a complete square , which is never negative, and four units are added.

So, it is the parabola with the minimum in 4 units above the x-axis.
Thus it never becomes equal to zero.


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