How do you find the zeros of # y = 3x^2 - 13x +15 # using the quadratic formula?

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Answer 1 · 2017-08-15T18:23:28

This sucker has complex roots!

#x = (-B +- sqrt((B^2-4AC)))/(2A) #

Plugging in A = 3, B = -13, C = 15 gives you:

#(13 +- sqrt(169 - 180))/6#

or #(13 +- sqrt(-11))/6#

Which is your problem, since there is no square root of a negative number.

Best you can do is rewrite it as a COMPLEX number of form #a + bi# where i is taken to be #sqrt(-1)#.

So you can do a little rewrite of #sqrt(-11)# as #sqrt(-1 * 11)# = #sqrt(11) * i#

And then evaluate it once for the positive case of the #+-#, and once for the negative. Which gives roots of:

#13/6 + (sqrt(11)/6 * i)# = 2.1667 + 0.553i

And

#13/6 - (sqrt(11)/6 * i)# = 2.1667 - 0.553i

(Note that I've rounded off.