solve this math question
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For this equation,
x+2 >= 0 and 8 - x*x >= 0
x >= -2 and 8 >= x*x
So 2sqrt(2) >= x >= -2
but wait
[tex] \sqrt{x+2} \ \textgreater \ \sqrt{8-x^2} \\ x+2 \ \textgreater \ 8 - x^{2} \\ x^{2} + x - 6 \ \textgreater \ 0 \\ (x+3)(x-2) \ \textgreater \ 0 \\ x \ \textgreater \ 2[/tex]
finally
2sqrt(2) >= x > 2
x+2 >= 0 and 8 - x*x >= 0
x >= -2 and 8 >= x*x
So 2sqrt(2) >= x >= -2
but wait
[tex] \sqrt{x+2} \ \textgreater \ \sqrt{8-x^2} \\ x+2 \ \textgreater \ 8 - x^{2} \\ x^{2} + x - 6 \ \textgreater \ 0 \\ (x+3)(x-2) \ \textgreater \ 0 \\ x \ \textgreater \ 2[/tex]
finally
2sqrt(2) >= x > 2
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