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!

Algebra ->  Graphs -> 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!      Log On


   

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(20065) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, it's about real and complex numbers.

x%5E2+%2B+4x+%2B+8+=+0

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

y=x%5E2+%2B+4x+%2B+8

For the x-intercepts are when y = 0.  But as you see
when looking at the graph of y=x%5E2+%2B+4x+%2B+8

graph%28800%2F3%2C400%2C-6%2C2%2C-2%2C10%2Cx%5E2+%2B+4x+%2B+8%29 

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:

x%5E2+%2B+4x+%2B+8+=+0

a=1, b=4, c=8

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-%284%29+%2B-+sqrt%28+%284%29%5E2-4%2A%281%29%2A%288%29+%29%29%2F%282%2A%281%29%29+

x+=+%28-4+%2B-+sqrt%2816-32+%29%29%2F2+

x+=+%28-4+%2B-+sqrt%28-16+%29%29%2F2+

x+=+%28-4+%2B-+4i%29%2F2+

x+=+%284%28-1+%2B-+i%29%29%2F2+

x+=+2%28-1+%2B-+i%29+

x+=+-2+%2B-+2i+

The solutions are complex numbers, and do not 
appear on the graph of y=x%5E2+%2B+4x+%2B+8

Edwin

Answer by richard1234(7193) About Me  (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(52921) About Me  (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

x%5E2+%2B+4x+%2B+8 = %28x%2B2%29%5E2+%2B+4.

You have a complete square %28x%2B2%29%5E2, 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.