How do you solve #3s^2-26x+2=5s^2+1# by completing the square?

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Answer 1 · 2018-03-03T13:32:09

#color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)#

#3x^2 - 26x+ 2 = 5x^2 + 1#

#5x^2 - 3x^2 + 26x = 1#

#2(x^2 + (2 * x * (13/2)) = 1#

#x^2 + 2x * (13/2) = 1/2#

Adding # (13/2)^2 # to both sides,

#x^2 + 2x (13/2) + (13/2)^2 = 1/2 + (13/2)^2 = (171/4)#

#(x + 13/2)^2 = (sqrt(171/4))^2#

#x + 13/2 = +- sqrt(171/4)#

#x = +- sqrt(171/4) - 13/2#

#color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)#