How do you find the roots, real and imaginary, of #y= x^2 - 6x - (6x+1)^2 # using the quadratic formula?

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Answer 1 · 2017-02-22T06:44:50

We have real zeros #8/35+sqrt46/35# and #8/35-sqrt46/35#

The roots of a quadratic function #y=f(x)=ax^2+bx+c=0# are given by #(-b+-sqrt(b^2-4ac))/(2a)#

Here we have #y=f(x)=x^2-6x-(6x+1)^2#

= #x^2-6x-(36x^2+12x+1)#

= #x^2-6x-36x^2-12x-1#

= #-35x^2-18x-1#

Hence its zeros are #(-(-18)+-sqrt((-18)^2-4xx(-35)xx(-1)))/(2xx(-35))#

= #(18+-sqrt(324-140))/(-70)#

= #(18+-sqrt184)/(-70)#

= #(18+-2sqrt46)/(-70)#

i.e #8/35+-sqrt46/35#

i.e. #8/35+sqrt46/35# and #8/35-sqrt46/35#