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 $8/35-sqrt46/35$

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