How do you solve #3/4x^2-2x+1/2=0# using the quadratic formula?

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Answer 1 · 2015-08-25T14:45:46

Multiply through by #4# first to get integer coefficients, then use the quadratic formula to get:

#x = (4+-sqrt(10))/3#

You can just plug the coefficients straight into the formula, but it's probably tidier to multiply through by #4# first to get integer coefficients.

Multiply #3/4x^2-2x+1/2=0# through by #4# to get:

#3x^2-8x+2 = 0#

This is of the form #ax^2+bx+c = 0#, with #a=3#, #b=-8# and #c=2#.

Then using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)=(8+-sqrt(8^2-4*3*2))/(2*3)#

#= (8+-sqrt(64-24))/6 = (8+-sqrt(40))/6 = (8+-2sqrt(10))/6 = (4+-sqrt(10))/3#