How do you solve #x+sqrt(-2x^2+9)=3# and identify any restrictions?

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Answer 1 · 2018-04-18T04:18:25

#x=0, 2, -(3sqrt2)/2<=x<=(3sqrt2)/2#

Restrictions: #-2x^2+9>=0#

#=>-(3sqrt2)/2<=x<=(3sqrt2)/2#

#x+sqrt(-2x^2+9)=3#

#sqrt(-2x^2+9)=3-x# subtract #x# from both sides

#-2x^2+9=9-6x+x^2# square both sides

#3x^2-6x=0#

#3x(x-2)=0# solve quadratic equation

#x=0,x=2#