how do i find 64-x^2<0

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64-x^2=0 (x-8)(x+8) x=8     x=-8

1. Subtract 64 from both sides -x^2 < -64 It really doesn't make a difference if the negative is in front of x. When you square a negative, it becomes a positive. 2. Cancel the negative in front of the x x^2 < 64 3. Finally, take the square root of of x and 64 x < 8