Question 1177686
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1)  From one side, if  x^2 - 9 >= 0,  i.e.  {x <= -3  OR  x >= 3},


    then  |x^2 - 9| = x^2 - 9 > 7,  x^2 > 7+9 = 16,  which implies  (x < -4} OR {x > 4}.


        So, one set of solution is  (-oo,-4) U (4,oo).





2)  From the other side, if  x^2 - 9 < 0,  i.e.  {-3 < x < 3},


    then  |x^2 - 9| = 9-x^2 > 7,  9 - 7 > x^2,  x^2 < 2,  which implies  -sqrt(2) < x < sqrt(2).


        So, the other set of solution is  {{{-sqrt(2)}}} < x < {{{sqrt(2)}}}.




3)  Finally, the solution set is the union of three intervals


        (-oo,-4) U ( {{{-sqrt(2)}}}, {{{sqrt(2)}}} ) U (4,oo).




4)  Visually, the solution set is seen and confirmed from this plot



    {{{graph( 400, 400, -6, 6, -5, 15,        
              abs(x^2-9), 7

)}}}


       Plot  y = |x^2 - 9|  (red)  and  y = 7  (green)



The solution set is where the red line is ABOVE the green line.
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Solved.