How do you find the zeros, real and imaginary, of #y=- x^2-22x+6 # using the quadratic formula?

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Answer 1 · 2016-04-10T10:59:06

Zeros are #-11+sqrt127# and #-11-sqrt127#

The roots of general form of equation #ax^2+bx+c=0# are zeros of the general form of equation #y=ax^2+bx+c#.

In the given equation #y=-x^2-22x+6#, we see that #a=-1#, #b=-22# and #c=6#.

As discriminant #b^2-4ac=(-22)^2-4(-1)(6)=484+24=508#

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#,

the zeros are the given equation are #x=(-(-22)+-sqrt508)/(2*(-1))#

or #x=-(22+-sqrt508)/2=-(22+-2sqrt127)/2# i.e.

zeros are #-11+sqrt127# and #-11-sqrt127#