VeritasPrepKarishma wrote:
rs1 wrote:
Hi all,
I am trying to figure out why a "certain" approach to this question doesn't work, thus forcing me to doubt some of my basic math knowledge. Advice would be greatly appreciated:
The question:
If \sqrt{2x} + 1 = \sqrt{x+1} and x is not a positive integer, x = ?
I can simplify to:
x + 2 \sqrt{2x} = 0.
I understand that the easiest method would be to move 'x' from the LHS to the RHS. Square. Then solve. However, I am trying to figure out whether this could be solved 'without' moving 'x' to the LHS to the RHS. I am trying to understand this because this is the approach that I first took when I first solved this question and got into quite a mess (and a non-answer); so I'm trying to figure out where I went wrong with the method that I took..
Thanks
Is the equation this:
\(\sqrt{2x} + 1 = \sqrt{x+1}\)?
If yes, after you get
\(x + 2 \sqrt{2x} = 0\)
Take \(\sqrt{x}\) common:
\(\sqrt{x}*(\sqrt{x} + 2*\sqrt{2}) = 0\)
Since \(\sqrt{x}\) is non negative, \((\sqrt{x} + 2*\sqrt{2})\) must be positive.
So, \(\sqrt{x} = 0\)
x = 0
Hi Karishma,
Thanks for this! Your solution makes perfect sense. I had totally forgotten that I could take out the common term because the radical threw me off.
The solutions to this equation are x=0 or x=8. Typically, I would solve such an equation by equating each term to zero:
\(\sqrt{x} = 0\)
or
\((\sqrt{x} + 2*\sqrt{2})=0\)
How would you solve the second equation: \((\sqrt{x} + 2*\sqrt{2})=0\) without moving any term to the RHS? I run into the same issues as my initial problem from this point. Is it possible to square both sides such that the LHS becomes an (a+b)^2 pattern? Or must I take out a common term in this scenario (as you did before?)
Feedback is appreciated!