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

1 Answer
May 25, 2018

Answer:

#x = 7/18+-sqrt(157)/18#

Explanation:

Given:

#y = 9x^2-7x-3#

Note that this is in standard form:

#y = ax^2+bx+c#

with #a=9#, #b=-7# and #c=-3#

It has zeros which we can find using the quadratic formula:

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

#color(white)(x) = (-(color(blue)(-7))+-sqrt((color(blue)(-7))^2-4(color(blue)(9))(color(blue)(-3))))/(2(color(blue)(9)))#

#color(white)(x) = (7+-sqrt(49+108))/(18)#

#color(white)(x) = (7+-sqrt(157))/(18)#

#color(white)(x) = 7/18+-sqrt(157)/18#

Note that #157# is prime, so #sqrt(157)# is already in simplest form.