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rs1
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If (root 2x) + 1 = (root x + 1), x = ? Question about approach.. 22 Mar 2016, 09:08
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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 (root 2x) + 1 = (root x + 1) and x is not a positive integer, x = ?

I can simplify to:

x + 2 (root 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' from LHS to RHS? I am trying to understand this because I didn't move x to the RHS during my first attempt when solving this question; I 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 (albeit the less efficient one)..

Thanks
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KarishmaB
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If (root 2x) + 1 = (root x + 1), x = ? Question about approach.. 22 Mar 2016, 09:16
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rs1
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
Show more

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
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rs1
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If (root 2x) + 1 = (root x + 1), x = ? Question about approach.. 22 Mar 2016, 11:18
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VeritasPrepKarishma
rs1
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
Show more


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!
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KarishmaB
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If (root 2x) + 1 = (root x + 1), x = ? Question about approach.. 28 Mar 2016, 08:37
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rs1
VeritasPrepKarishma
rs1
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!
Show more

Say if the equation you get is
\(5\sqrt{x} = 0\)
what would you say is the value of x? You would simply say that x = 0, right?

Similarly, here \(\sqrt{x}*(\sqrt{x} + 2*\sqrt{2})=0\)
you need to note that \((\sqrt{x} + 2*\sqrt{2})\) is a positive number so it cannot be 0. A square root is always non negative so a square root + a positive number must be a positive number only. So you cannot equate it to 0.
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If (root 2x) + 1 = (root x + 1), x = ? Question about approach.. 31 Mar 2016, 01:28
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Quote:

could you please show me how you simplified the given equation into "x + 2?2x = 0"
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\(\sqrt{2x} + 1 = \sqrt{x+1}\)

Square both sides:
\((\sqrt{2x})^2 + 1^2 + 2*\sqrt{2x}*1 = x + 1\)

\(2x + 1 + 2*\sqrt{2x} = x + 1\)

\(x + 2*\sqrt{2x} = 0\) ...........(I)

We know that \(\sqrt{x}*\sqrt{x} = x\), so substituting in (I)

\(\sqrt{x}*\sqrt{x} + 2*\sqrt{2x} = 0\)

Take \(\sqrt{x}\) common,

\(\sqrt{x} ( \sqrt{x} + 2\sqrt{2}) = 0\)