How do you find the zeros, real and imaginary, of #y=- x^2+12x+6 # using the quadratic formula?

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Answer 1 · 2018-01-13T12:30:14

Solution ; Zeros are two real number #x ~~12.6 , x ~~ -0.48 #

#y=-x^2+12x+6 # Comparing with standard quadratic

equation #ax^2 + bx + c = 0# and #a# is not zero. Here

#a=-1 . b= 12 ,c = 6 ; D# is discriminant and #D=b^2-4ac#

#:.D=b^2-4ac= 12^2-4*(-1)*6= 168# . If #D# is positive,

we get two real solutions, if it is zero just one solution, and

if it is negative we get two complex (imaginary) solutions.

Quadratic Formula: #x = [ -b +- sqrt((b^2-4ac)) ] / (2a)#

#:. x = [ -12 +- sqrt(12^2-4*(-1)* 6)] / ( 2 *(-1)) # or

# x = [ -12 +- sqrt(168)] / (-2) ~~ 6 +- 6.48 # or

#x ~~12.48 (2dp) and x ~~ -0.48 (2dp)#

Solution ; Zeros are two real number #x ~~12.6 , x ~~ -0.48 #

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