Question

How do you find the roots, real and imaginary, of `y=-5x^2 +(x -4)^2` using the quadratic formula?

Answer 1

We have two real zeros, which are irrational. These are `-1+-sqrt5`

Explanation:

Quadratic formula uses general form of quadratic equation `ax^2+bx+c=0` abd gives the solution of this equation as `x=(-b+-sqrt(b^2-4ac))/(2a)`. If `alpha` and`beta` are roots of this equation, then they are also zeros of the function `y=ax^2+bx+c`.

Hence, we should first convert the function `y=-5x^2+(x-4)^2`, to general form. Simplifying it we get,

`y=-5x^2+x^2-8x+16`

or `y=-4x^2-8x+16` and hence zeros of this function are

`x=(-(-8)+-sqrt((-8)^2-4*(-4)*16))/(2*(-4))`

or `x=(8+-sqrt(64+256))/(-8)=(8+-sqrt(320))/(-8)`

or `x=(8+-8sqrt5)/(-8)=-1+-sqrt5`

Hence, we have two real zeros, which are irrational.