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

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Answer 1 · 2018-06-02T22:07:52

x=3/4

#ax^2+bx+c#

#4/3x^2 - 2x + 3/4 = 0#

I would start by multiplying the whole thing by the common denominator of the fractions: #12# then we can just deal with integers.

#12(4/3x^2 - 2x + 3/4 = 0)#

#16x^2 -24x +9 =0#

#a=16, b=-24, c=9#

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

#x=(-(-24)+-sqrt((-24)^2-4*16*9))/(2*16)#

#x=(24+-sqrt(576-576))/(32)#

#x=24/32=3/4#

graph{16x^2 -24x +9 [-9.54, 10.46, -1.2, 8.8]}