How do you find the zeros, real and imaginary, of #y=x^2+x-4# using the quadratic formula?

1 Answer
May 15, 2016

Answer:

#x =(-1+-sqrt(17))/2#

Explanation:

#x^2+x-4# is of the form #ax^2+bx+c# with #a=1#, #b=1# and #c=-4#

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-1+-sqrt(1^2-(4*1*(-4))))/(2*1)#

#=(-1+-sqrt(1+16))/2#

#=(-1+-sqrt(17))/2#

Footnote

Notice the expression #+-sqrt(b^2-4ac)# in the formula.

The radicand #b^2-4ac# is called the discriminant and often represented by the capital Greek letter delta #Delta#.

By examining whether #Delta > 0#, #Delta = 0# or #Delta < 0# we can tell what kind of zeros a quadratic (with Real coefficients) has:

  • 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 non-Real Complex zeros which are Complex conjugates of one another.