How do you find the zeros, real and imaginary, of #y=x^2+x4# using the quadratic formula?
1 Answer
Explanation:
This has zeros given by the quadratic formula:
#x = (b+sqrt(b^24ac))/(2a)#
#=(1+sqrt(1^2(4*1*(4))))/(2*1)#
#=(1+sqrt(1+16))/2#
#=(1+sqrt(17))/2#
Footnote
Notice the expression
The radicand
By examining whether

If
#Delta > 0# then the quadratic has two distinct Real zeros. In addition, if#Delta# is a perfect square, then the zeros of the quadratic are rational (assuming the coefficients of the quadaratic are). 
If
#Delta = 0# then the quadratic has one repeated Real zero. 
If
#Delta < 0# then the quadratic has two nonReal Complex zeros which are Complex conjugates of one another.