Question

How do you find the zeros, real and imaginary, of `y=x^2/3+1/7x-5/9` using the quadratic formula?

Answer 1

`x=(-3+-sqrt(1007/3))/14`

Explanation:

The quadratic formula states that the zeros of a quadratic function in the form `y=ax^2+bx+c` are equal to

`x=(-b+-sqrt(b^2-4ac))/(2a)`

With the quadratic `y=x^2/3+1/7x-5/9`, we can see that

`{(a=1/3),(b=1/7),(c=-5/9):}`

Plug these into the quadratic formula:

`x=(-1/7+-sqrt((1/7)^2-(4xx1/3xx-5/9)))/(2xx1/3)`

Simplify.

`x=(-1/7+-sqrt(1/49+20/27))/(2/3)`

`x=(-1/7+-sqrt(27/1323+980/1323))/(2/3)`

`x=(-1/7+-sqrt(1007/1323))/(2/3)`

`x=(-1/7+-sqrt(1007/(9xx49xx3)))/(2/3)`

`x=(-1/7+-1/21sqrt(1007/3))/(2/3)`

`x=(-1/7+-1/21sqrt(1007/3))*(3/2)`

`x=(-3/7+-1/7sqrt(1007/3))/2`

`x=(-3+-sqrt(1007/3))/14`