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

1 Answer
Jan 25, 2016

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