How do you find the zeros, if any, of #y= 3x^2 -2x+ 8 #using the quadratic formula?

1 Answer
Feb 12, 2017

Answer:

There are only complex zeros: #x = 1/3 +- sqrt(23)/3 i#

Explanation:

Use the quadratic formula where #A = 3, B = -2, C = 8#

#x =( -(-2)+- sqrt((-2)^2-4(3)(8))/(2(3))) = ((2+-sqrt(-92))/6)#

#sqrt(-92) = sqrt(4)sqrt(-23) = 2sqrt(23)i#

Therefore #x = 1/3 +- sqrt(23)/3 i#

If you graph the function it does not cross the x-axis: graph{3x^2-2x+8 [-32.23, 31.8, -5.5, 26.55]}