Question

How do you find the zeros, real and imaginary, of `y= 3x^2-43x-73` using the quadratic formula?

Answer 1

The zeros of the given Quadratic Polynomial are `(43+-sqrt(2725))/6.`

Explanation:

Formula to find the zeros, say `alpha` & `beta,` of a Quadratic Polynomial `y=P(x)=ax^2+bx+c` is

`alpha,beta=(-b+-sqrt(b^2-4ac))/(2a)`

In our case, `a=3, b=-43,c=-73.`
`:. alpha,beta= {43+-sqrt(1849+876)}/6=(43+-sqrt(2725))/6.`

We note that since `Delta=b^2-4ac=2725>0`, the given quadratic can not have any complex zero.