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

1 Answer
Apr 18, 2018

Answer:

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

Explanation:

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#