How do you find the zeros of # y = -x^2 - 1/11x + 1/7 # using the quadratic formula?

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Answer 1 · 2016-08-03T07:39:35

Zeros are #7/22-7/2sqrt(491/847)# and #7/22+7/2sqrt(491/847)#

Quadratic formula gives the zeros of #y=ax^2+bx+c# as #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Hence zeros of #y=-x^2-1/11x+1/7# are

#x=(-(-1/11)+-sqrt((-1/11)^2-4(-1)(1/7)))/(2(1/7))#

= #((1/11)+-sqrt(1/121+4/7))/(2/7)#

= #(1/11+-sqrt((7+484)/847))xx7/2#

= #7/22+-7/2sqrt(491/847)#

Hence zeros are #7/22-7/2sqrt(491/847)# and #7/22+7/2sqrt(491/847)#