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August 30, 2014

August 30, 2014

Posted by **sandy** on Saturday, October 20, 2012 at 9:20am.

Thanks

- college alegebra -
**Reiny**, Saturday, October 20, 2012 at 10:28amI will assume you mean

√(3x-2) + √(2x+5) + 1 = 0

by definition, √(anything) is the POSITIVE square root of anything, so the left side of your equation is the sum of three positive numbers, which of course can never be zero.

your equation has no solution.

why don't you graph

y = √(3x-2) + √(2x+5) + 1

and you will see that the entire graph lies above the x-axis.

Here is what Wolfram says:

http://www.wolframalpha.com/input/?i=√%283x-2%29+%2B+√%282x%2B5%29+%2B+1+

the "real number" graph is in blue, lies above the x-axis

- college alegebra -
**Shell**, Sunday, October 21, 2012 at 11:45amReiny, no, actually, the square root of anything is not always the positive. because if you multiply two negative numbers, you get a positive number. So, when you square root anything, you assume it is either the positive or negative of a number, because it can be either.

Unfortunately, I do not have a TI-84, so I can only say to graph both the positive and the negative forms on you graph and see where they intersect.

Or you can solve the problem the old fashion way (which is rather easy...)

- college alegebra -
**Shell**, Sunday, October 21, 2012 at 11:47am....ok, maybe it's not easy... >.<

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