How do you find the zeros of # y = -x^2 - 9x + 3 # using the quadratic formula?

1 Answer
Jul 21, 2016

Answer:

#-9/2-sqrt93/2# and #-9/2+sqrt93/2#,

are real zeros of #y=-x^2-9x+3#.

Explanation:

Zeros of #y=-x^2-9x+3# are those values of #x# for which #y=0#. Hence, zeros are given by the solution of quadratic equation #-x^2-9x+3=0# or #x^2+9x-3=0#.

Quadratic formula gives solution of equation #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#. Hence for #x^2+9x-3=0#, we have

#x=(-9+-sqrt(9^2-4*1*(-3)))/(2*1)#

= #(-9+-sqrt(81+12))/2#

= #(-9+-sqrt93)/2#

= #-9/2-sqrt93/2# and #-9/2+sqrt93/2#,

are real zeros of #y=-x^2-9x+3#.