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 $x=13/2+sqrt165/2$

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