What is x if #3x - sqrt(2(x-3)^2) = 2x + 7-5x#?

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Answer 1 · 2018-08-04T12:52:17

#x_1=(72+22sqrt2)/(34*2)=(36+11sqrt2)/34# and #x_2=(72-22sqrt2)/(34*2)=(36-11sqrt2)/34#

#3x-sqrt(2*(x-3)^2)=2x+7-5x#

#3x-sqrt(2*(x-3)^2)=7-3x#

#3x+3x-7=sqrt(2*(x-3)^2)#

#6x-7=sqrt(2*(x-3)^2)#

#(6x-7)^2=2*(x-3)^2#

#36x^2-84x+49=2*(x^2-6x+9)#

#36x^2-84x+49=2x^2-12x+18#

#34x^2-72x+31=0#

#Delta=(-72)^2-4*34*31=968=(22sqrt2)^2#

Hence #x_1=(72+22sqrt2)/(34*2)=(36+11sqrt2)/34# and #x_2=(72-22sqrt2)/(34*2)=(36-11sqrt2)/34#