How do you find the zeros, real and imaginary, of #y=-x^2- x -35# using the quadratic formula?

1 Answer
May 30, 2016

Answer:

The zeros are #x=-0.5+i5.895# and #x=-0.5-i5.895#.

Explanation:

The zeros are the solution of this equations

#-x^2-x-35=0#

first of all we can remove all the minus multiplying left and right for #-1#, but on the right there is zero that does not change sign.

#x^2+x+35=0#

The quadratic formula tells us that an equation in the form
#ax^2+bX+c=0# has solutions #x=(-b\pmsqrt(b^2-4ac))/(2a)#.

For us #a=1#, #b=1# and #c=35#, I substitute

#x=(-1\pmsqrt(1^2-4*1*35))/(2)#
#=(-1\pmsqrt(-139))/2#
We notice that the root is negative and then imaginary. So I rewrite it using the fact that #i=sqrt(-1)#

#x=(-1\pmisqrt(139))/2\approx-1/2\pmi11.79/2=-0.5\pmi5.895#

The two solutions are then #x=-0.5+i5.895# and #x=-0.5-i5.895#.