Question 131657
SOLUTION -
  f(x) = {{{3x+3sqrt(x)-6}}}
       taking {{{sqrt (x)=y}}}
        therefore {{{x=y^2}}}
we have
    {{{3y^2+3y-6}}}
    {{{3y^2+6y-3y-6}}}
    {{{3y(y+2)-3(y+2)}}}
    {{{(y+2)(3y-3)}}}
     {{{y+2=0}}}
     {{{y=-2}}}
    {{{3y-3=0}}}
    {{{3y=3}}}
    {{{y=1}}}

now {{{sqrt (x)=y}}}
substituting the value of y we have
     when y=-2
     {{{sqrt (x)=-2}}}
     {{{x=(-2)^2}}}
      {{{x=4}}}
      when y=1
     {{{sqrt (x)=1}}}
     {{{x=1^2}}}
        x=1

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