How do you solve #3x^2-4x+9=-5# by completing the square?

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Answer 1 · 2017-03-16T23:34:08

This has no real solution. Proven below, and by the fact that the graph of #f(x)=3x^2-4x+14# is always positive.
graph{3x^2-4x+14 [-20,20, -2, 20]}

#3x^2-4x+9 = -5#

#3x^2-4x+14=0#

#3[x^2-4/3x+14/3]=0#

#3[(x-2/3)^2+38/9]=0#

#(x-2/3)^2+38/9=0#

#(x-2/3)^2=-38/9#

#x-2/3=+-sqrt(-38/9)#

#x=2/3+-sqrt(-38/9)#

#x=frac{2+-isqrt38}{3}#