Posted by Sarah on Saturday, January 8, 2011 at 4:48pm.
I have to solve sqrt(2x7) >=5
answers would be x>=16
x>=7/2
7/2<=x<=16
x<=16
I don't get this concept at all and I have 20 of these kinds of problems for homeowrkcan someone direct me how to start or give me an example or just help please? I really need to figure out this concept
Thank you

AlgebraI really need help, please?  Reiny, Saturday, January 8, 2011 at 5:04pm
√(2x7) > 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
2x7 ≥ 25
2x ≥ 32
x ≥ 16
and of course for √(2x7) to be a real number, the inside must be ≥ 0
that is, 2x7 ≥ 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. )

AlgebraI really need help, please?  Sarah, Saturday, January 8, 2011 at 5:07pm
Thank you for the detailed exampleI can really use those to figure out the rest of my homeowrkI should be good now
Thanks again

AlgebraI really need help, please?  Reiny, Saturday, January 8, 2011 at 5:09pm
welcome, come back if you run into more problems.

AlgebraI really need help, please?  helper, Saturday, January 8, 2011 at 5:11pm
sqrt(2x7) >= 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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