SOLUTION: Find the x-intercepts. y = x^2 + 4

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Question 173629This question is from textbook Introductory Algebra
: Find the x-intercepts.
y = x^2 + 4 This question is from textbook Introductory Algebra

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+x%5E2+%2B+4
to find the x-intercepts, you want to find the values for x that result in y=0
0+=+x%5E2+%2B+4
You can plot this or use the quadratic equation to solve
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B4+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A1%2A4=-16.

The discriminant -16 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -16 is + or - sqrt%28+16%29+=+4.

The solution is x%5B12%5D+=+%28-0%2B-+i%2Asqrt%28+-16+%29%29%2F2%5C1+=++%28-0%2B-+i%2A4%29%2F2%5C1+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B4+%29