# Math Help - Solving for X

1. ## Solving for X

Squareroot of x equals -1

2. \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

3. Originally Posted by Math1337
Squareroot of x equals -1
$\sqrt{x}=-1$
So...
$(\sqrt{x})^2=(-1)^2$

See if that leads you anywhere.

4. I have determined that no real solution exists. I susppect that the answer involves I added to somthing

5. The full problem is sqrroot Of 3 - (inside the square root) the squareroot of x plus 5. All equals 2

6. Just to clarify. Do you mean that the original problem is to solve for x in the equation

$\sqrt{3-\sqrt{x+5}}=2?$

7. Originally Posted by Ackbeet
Just to clarify. Do you mean that the original problem is to solve for x in the equation

$\sqrt{3-\sqrt{x+5}}=2?$
Yes

8. 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.

9. 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.

10. 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.