Algebra-I really need help, please?

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I have to solve sqrt(2x-7) >=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 homeowrk-can 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

  • Algebra-I really need help, please? -

    √(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. )

  • Algebra-I really need help, please? -

    Thank you for the detailed example-I can really use those to figure out the rest of my homeowrk-I should be good now

    Thanks again

  • Algebra-I really need help, please? -

    welcome, come back if you run into more problems.

  • Algebra-I really need help, please? -

    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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