How do you find the roots, real and imaginary, of #y= 3x^2-5x- 1 # using the quadratic formula?

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Answer 1 · 2016-01-29T08:02:24

#x=5/6+(sqrt3)/2,# #5/6-(sqrt 3)/2#

#y=3x^2-5x-1# is a quadratic equation in standard form, #y=ax+bx+c#, where #a=3, b=-5, c=-1#, and #y=0#.

The roots are the solutions for #x#.

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

Substitute the known values into the equation.

#x=(-(-5)+-sqrt((-5)^2-(4*3*-1)))/(2*3)#

Simplify.

#x=(5+-sqrt(25+12))/6#

Add #25# and #12#.

#x=(5+-sqrt27)/6#

Simplify #sqrt(27)#.

#sqrt27=sqrt(3*3*3)=sqrt(3^2*3)=3sqrt 3#

#x=(5+-3sqrt3)/6#

Write the value for #x# as two fractions.

#x=5/6+-(3sqrt3)/6#

Simplify #3/6# to #1/2#.

#x=5/6+-(sqrt3)/2#