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This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

Hey, below mentioned is my strategy to solve this problem,

Sqrt both side you will get, 3-2x= 2x+1+2sqrt{2x} 3-2x-2x-1= 2sqrt{2x} 2-4x= 2sqrt{2x} devide by 2 both side 1-2x = sqrt{2x} Again sqrt of both side 1+4x^2-4x= 2x 1+4x^2-6x=0 4x^2= 6x-1

This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

Hey, below mentioned is my strategy to solve this problem,

Sqrt both side you will get, 3-2x= 2x+1+2sqrt{2x} 3-2x-2x-1= 2sqrt{2x} 2-4x= 2sqrt{2x} devide by 2 both side 1-2x = sqrt{2x} Again sqrt of both side 1+4x^2-4x= 2x 1+4x^2-6x=0 4x^2= 6x-1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) --> \(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side: \(2-4x=2*\sqrt{2x}\) --> reduce by 2: \(1-2x=\sqrt{2x}\) --> square again: \((1-2x)^2=(\sqrt{2x})^2\) --> \(1-4x+4x^2=2x\) --> rearrange again: \(4x^2=6x-1\).

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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26 Jun 2012, 03:37

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Expert's post

metallicafan wrote:

Solving the equation, I arrive to this point: \(2 - 2\sqrt{2x}\)\(=\)\(4x\)

How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution?

After this point, you see that none of the options has a square root sign. You need to get rid of it to get to the answer. Hence, it is apparent that you need to square it again. You then take the square root term to one side and the rest of the terms to the other side and square both sides of the equation. This helps you get rid of the square root sign. _________________

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