SOLUTION: How would I solve this using quadratic formula? f(x) = x^2 + 4

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Question 139521: How would I solve this using quadratic formula?
f(x) = x^2 + 4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

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



So lets solve x%5E2%2B4=0 (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like x%5E2%2B0%2Ax%2B4=0 notice a=1, b=0, and c=4)




x+=+%280+%2B-+sqrt%28+%280%29%5E2-4%2A1%2A4+%29%29%2F%282%2A1%29 Plug in a=1, b=0, and c=4



x+=+%280+%2B-+sqrt%28+0-4%2A1%2A4+%29%29%2F%282%2A1%29 Square 0 to get 0



x+=+%280+%2B-+sqrt%28+0%2B-16+%29%29%2F%282%2A1%29 Multiply -4%2A4%2A1 to get -16



x+=+%280+%2B-+sqrt%28+-16+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%280+%2B-+4%2Ai%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%280+%2B-+4%2Ai%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=0+%2B+2%2Ai or x=0+-+2%2Ai

Notice if we graph the quadratic y=x%5E2%2B4, we get

+graph%28+500%2C+500%2C+-15%2C+15%2C+-11%2C+19%2C+x%5E2%2B4%29+ graph of y=x%5E2%2B4

And we can see that there are no real roots

To visually verify the answer, check out this page to see a visual representation of imaginary roots