How do you find the roots, real and imaginary, of #y=8x^2+7x-(3x-1)^2 # using the quadratic formula?

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Answer 1 · 2016-04-10T11:16:54

Zeros of function are #x=13/2-sqrt165/2# and #x=13/2+sqrt165/2#

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#.

Simplifying the given equation #y=8x^2+7x-(3x-1)^2#, gives us

#y=8x^2+7x-(9x^2-6x+1)=-x^2+13x-1#

we see that #a=-1#, #b=13# and #c=-1#.

As discriminant #b^2-4ac=(13)^2-4(-1)(-1)=169-4=165#

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=(-(13)+-sqrt165)/(2*(-1))#

or Zeros of function are #x=13/2-sqrt165/2# and #x=13/2+sqrt165/2#