√(2x-7) > 5
Since both left and right side are positive we don't have to worry about any changes in direction of the inequality sign.
So square both sides
2x-7 ≥ 25
2x ≥ 32
x ≥ 16
and of course for √(2x-7) to be a real number, the inside must be ≥ 0
that is, 2x-7 ≥ 0
2x ≥ 7
x ≥ 7/2
so now you have x ≥ 7/2 AND x ≥ 16
which set of numbers satisfies BOTH conditions?
It would be x ≥ 16
(take x = 5 for an example
it satisfies the first condition, in that it is ≥ 7/2
but does not work for the original inequation. )
Thank you for the detailed example-I can really use those to figure out the rest of my homeowrk-I should be good now
welcome, come back if you run into more problems.
sqrt(2x-7) >= 5
What don't you understand?
to solve, square both sides to get rid of the radical
2x - 7 >= 25
2x >= 32
x >= 16
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