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If root{3-2x} = root(2x) +1, then 4x^2 =

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

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New post 30 May 2015, 03:43
reto wrote:
Bunuel wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600


Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty?

If you square the expression \(\sqrt{2x} +1\) and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?` :roll:

Thank you so much


1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question.
2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 30 May 2015, 04:04
Bunuel wrote:
reto wrote:
Bunuel wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600


Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty?

If you square the expression \(\sqrt{2x} +1\) and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?` :roll:

Thank you so much


1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question.
2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually.


Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis - why? I know it's not the same. But I don't understand why you completed it when squaring...
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 30 May 2015, 04:21
1
reto wrote:
Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis - why? I know it's not the same. But I don't understand why you completed it when squaring...


Squaring means multiplying by the same number, so when you square a + b, you get (a + b)(a + b) = (a + b)^2.

I'd advice to brush up fundamentals before practicing questions, especially, hard ones.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 20 Mar 2016, 17:25
Bunuel wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600


sqrt(3-2x)-1 = sqrt(2x) | square everything
2x = 3-2x - 2*sqrt(3-2x)+1
2x=4 - 2x - 2*sqrt(3-2x)
4-4x=2*sqrt(3-2x) | square everything again
16+16x^2 -32x = 12 - 8x | add 8x to both sides, subtract 12 from both sides
16x^2 - 24x +4 = 0 | divide by 4
4x^2 -6x +1 = 0 | add 6x to both sides and subtract 1:
4x^2 = 6x-1

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

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New post 08 Mar 2018, 17:48
Bunuel wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600

Main Idea:While simplifying keep the more complex root expression alone on one side and simplify.

Details: sqrt(3-2x)=sqrt(2) +1.

Squaring on both sides: 3-2x=2x+1+2*sqrt(2x)

Simplifying we get (4x)^2 = 6x-1

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

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New post 02 May 2018, 07:51
Bunuel wrote:
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 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: \(4x^2=6x-1\).

Answer: E.




Bunuel can you please share your expertise knowledge on the following :)

how from this \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) we get this \(3-2x=2x+2*\sqrt{2x}+1\) ? :?

thank you for your kind attention to my question :)
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If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 02 May 2018, 08:17
dave13 wrote:
Bunuel wrote:
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 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: \(4x^2=6x-1\).

Answer: E.




Bunuel can you please share your expertise knowledge on the following :)

how from this \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) we get this \(3-2x=2x+2*\sqrt{2x}+1\) ? :?

thank you for your kind attention to my question :)


Let me asks you:

1. What does \((\sqrt{a})^2\) equal to?
2. What does \((a + b)^2\) equal to?

If you can answer these questions, you should be able to figure out your doubt.
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If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 02 May 2018, 09:36
Bunuel wrote:
dave13 wrote:
Bunuel wrote:
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 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: \(4x^2=6x-1\).

Answer: E.




Bunuel can you please share your expertise knowledge on the following :)

how from this \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) we get this \(3-2x=2x+2*\sqrt{2x}+1\) ? :?

thank you for your kind attention to my question :)


Let me asks you:

1. What does \((\sqrt{a})^2\) equal to?
2. What does \((a + b)^2\) equal to?

If you can answer these questions, you should be able to figure out your doubt.


Hi Bunuel

thanks for your questions, i will gladly answer your questions:)

1. What does \((\sqrt{a})^2\) equal to? it equals \(a\)

2. What does \((a + b)^2\) equal to? it equals this \((a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2\)

now let me get back to the question

\((\sqrt{3-2x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\)

so we get \((3-2x)^2 = (2x-1)^2\)

now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\)

we get \((3-2x) ( 3-2x) = (2x+1) (2x+1)\) so from here we get \(10=16x\):?

can you please explain what am i doing wrong :? knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes :)
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 02 May 2018, 11:28
1
dave13 wrote:
Bunuel wrote:
dave13 wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 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: \(4x^2=6x-1\).

Answer: E.


Bunuel can you please share your expertise knowledge on the following :)

how from this \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) we get this \(3-2x=2x+2*\sqrt{2x}+1\) ? :?

thank you for your kind attention to my question :)


Let me asks you:

1. What does \((\sqrt{a})^2\) equal to?
2. What does \((a + b)^2\) equal to?

If you can answer these questions, you should be able to figure out your doubt.


Hi Bunuel

thanks for your questions, i will gladly answer your questions:)

1. What does \((\sqrt{a})^2\) equal to? it equals \(a\)

2. What does \((a + b)^2\) equal to? it equals this \((a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2\)

now let me get back to the question

\((\sqrt{3-2x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\)

so we get \((3-2x)^2 = (2x-1)^2\)

now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\)

we get \((3-2x) ( 3-2x) = (2x+1) (2x+1)\) so from here we get \(10=16x\):?

can you please explain what am i doing wrong :? knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes :)


1. Yes, \((\sqrt{a})^2=a\), hence \((\sqrt{3-2x})^2 =3-2x\).

2. Here you are also correct: \((a + b)^2=a^2+2ab+b^2\), hence \((\sqrt{2x} +1)^2=(\sqrt{2x})^2 + 2*\sqrt{2x}*1+1^2=2x+2*\sqrt{2x}+1\).

Does this make sense?
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If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 03 May 2018, 09:31
Bunuel, it makes perfect sense ! :)

i applied this formula \((a - b)^2\) = \(a^2-2ab+b^2\) to \((1-2x)^2=(\sqrt{2x})^2\)

so i get this \(1^2 - 2 * 1 *(-2x) +2x^2 = 2x\) ----> \(1 -2-2x+4x=2x\) --> \(1 -2+2x= 2x\) ---> \(-1 =0\):? :) what am i doing wrong , can you please explain in two or three words :)

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

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New post 03 May 2018, 10:16
1
dave13 wrote:
Bunuel, it makes perfect sense ! :)

i applied this formula \((a - b)^2\) = \(a^2-2ab+b^2\) to \((1-2x)^2=(\sqrt{2x})^2\)

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> \(1 -2-2x+4x=2x\) --> \(1 -2+2x= 2x\) ---> \(-1 =0\):? :) what am i doing wrong , can you please explain in two or three words :)

many thanks! :-)


I highlighted the mistakes.

\((1 - 2x)^2 = 1 - 2*2x + (2x)^2 = 1 - 4x + 4x^2\)
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Collection of Questions:
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If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 03 May 2018, 11:19
Bunuel wrote:
dave13 wrote:
Bunuel, it makes perfect sense ! :)

i applied this formula \((a - b)^2\) = \(a^2-2ab+b^2\) to \((1-2x)^2=(\sqrt{2x})^2\)

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> \(1 -2-2x+4x=2x\) --> \(1 -2+2x= 2x\) ---> \(-1 =0\):? :) what am i doing wrong , can you please explain in two or three words :)

many thanks! :-)


I highlighted the mistakes.

\((1 - 2x)^2 = 1 - 2*2x + (2x)^2 = 1 - 4x + 4x^2\)



Bunuel thank you ! :) just last question :-)

how did you convert this \((2x)^2\) and got \(4x^2\) :?

\((2x)^2\) equals \(4x\) and not \(4x^2\)

please explain this part and I will be more than happy :)
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 03 May 2018, 11:21
1
dave13 wrote:
Bunuel wrote:
dave13 wrote:
Bunuel, it makes perfect sense ! :)

i applied this formula \((a - b)^2\) = \(a^2-2ab+b^2\) to \((1-2x)^2=(\sqrt{2x})^2\)

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> \(1 -2-2x+4x=2x\) --> \(1 -2+2x= 2x\) ---> \(-1 =0\):? :) what am i doing wrong , can you please explain in two or three words :)

many thanks! :-)


I highlighted the mistakes.

\((1 - 2x)^2 = 1 - 2*2x + (2x)^2 = 1 - 4x + 4x^2\)



Bunuel thank you ! :) just last question :-)

how did you convert this \((2x)^2\) and got \(4x^2\) :?

\((2x)^2\) equals \(4x\)and not \(4x^2\)

please explain this part and I will be more than happy :)


We have \((2x)^2\). Notice that we are squaring 2x, so you should square both 2 and x: \((2x)^2=2x*2x=4x^2\)
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Collection of Questions:
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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New post 17 Aug 2018, 15:21
Is there a particular thinking format or approach we should apply on such questions. ? I did arrive at solution but took almost 4 minutes. My initial thinking was isolating 4x^2 o one side but this could be done in more than 1 way. Is there something i can keep in mind while approaching such questions other than trial and error.
Thanks in advance.

Siddharth


Bunuel wrote:
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 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: \(4x^2=6x-1\).

Answer: E.
GMAT Club Bot
Re: If root{3-2x} = root(2x) +1, then 4x^2 = &nbs [#permalink] 17 Aug 2018, 15:21

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