Question

Inequality to x2 < 64

Answer 1

              NOT MY WORDS TAKEN FROM A SOURCE!

(x^2) <64  => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2    so    (x^2) - (8^2) < 0  To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y)  You can work backwards and verify this is true. so let's set (x^2) - (8^2)  equal to zero to find the roots: (x^2) - (8^2) = 0   => (x-8)*(x+8) = 0       if x-8 = 0 => x=8      and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0  so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So  -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.

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