Squareroot of x equals -1

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- May 26th 2011, 08:06 AMMath1337Solving for X
Squareroot of x equals -1

- May 26th 2011, 08:19 AMmathprofessor
\sqrt{x}can only equal 0 or positive numbers (if we are looking for real solutions), so your problem has no solution.

Whenever you square both sides of an equation you can get erroneous roots. For example, suppose x=4 (so x equals 4 and no other number!!). If we square both sides we get {x}^{2}= 16. Solving for x, we get x= \pm 4. We must reject our erroneous solution of x = -4.

So, if we use the above step and square both sides we get x = 1. A quick checks yields that \sqrt{1}=-1, that is 1 = -1 which of course is absurd implying that there is no real solution - May 26th 2011, 08:19 AMQuacky
- May 26th 2011, 08:22 AMMath1337
I have determined that no real solution exists. I susppect that the answer involves I added to somthing

- May 26th 2011, 08:25 AMMath1337
The full problem is sqrroot Of 3 - (inside the square root) the squareroot of x plus 5. All equals 2

- May 26th 2011, 08:58 AMAckbeet
Just to clarify. Do you mean that the original problem is to solve for x in the equation

- May 26th 2011, 09:10 AMMath1337
- May 26th 2011, 09:24 AMAckbeet
Right, so squaring both sides and bringing the 3 over to the RHS yields essentially the equation in the OP (y = x + 5 gets you there precisely, only in the variable y). I agree with the previous posters that there is no real solution. There are even issues with a complex solution. I think you'd have to define a different branch of the square root function in order for the equation to have a solution.

That's a complex analysis topic, though, so I don't think I'd worry about it for now, if you don't have to. - May 26th 2011, 09:34 AMMath1337
I actually was wanting to find the answer for a question on our take home test. We were allowed to use any source but nobody seems to be able to find it out. All we can see is it probably involves I.

- May 26th 2011, 09:37 AMAckbeet
I seriously doubt if "Using any source" means "Getting the answer from another person". Thread closed. PM me or a moderator if you wish to discuss it.