Question 139453
3+sqrt3x+1=x
sqrt3x=x-4
now square both sides:
3x=(x-4)^2
3x=x^2-8x+16
x^2-8x-3x+16=0
x^2-11x+16=0
usingthe quasratig equation:{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} we get.
x=(11+-sqrt[-11^2-4*1*16])/2*1
x=(11+-sqrt[121-64])/2
x=(11+-sqrt57)/2
x=(11+-7.55)/2
x=11+7.55)/2
x=18.55/2
x=9.275 answer.
x=(11-7.55)/2
x=3.45/2
x=1.725 answer.