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This can be solved by inspection. You can notice that the square of ANY real number is positive or zero, so "x" can be any real number. ("Greater than" and "less than" don't make much sense for complex numbers.)

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8y ago

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x^2 > -16Since x^2 >=0 for all real x, then the statement is true for all real x.

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8y ago
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lenpollock

Lvl 1
1y ago
Agreed!!!!

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Q: How do you solve X squared is greater than -16?
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