Posted by **Sarah** on Saturday, January 8, 2011 at 4:48pm.

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? -
**Reiny**, Saturday, January 8, 2011 at 5:04pm
√(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? -
**Sarah**, Saturday, January 8, 2011 at 5:07pm
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? -
**Reiny**, Saturday, January 8, 2011 at 5:09pm
welcome, come back if you run into more problems.

- Algebra-I really need help, please? -
**helper**, Saturday, January 8, 2011 at 5:11pm
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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