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Re: If root{32x} = root(2x) +1, then 4x^2 =
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30 May 2015, 03:43



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Re: If root{32x} = root(2x) +1, then 4x^2 =
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30 May 2015, 04:04
Bunuel wrote: reto wrote: Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), 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?` 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{32x} = root(2x) +1, then 4x^2 =
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30 May 2015, 04:21



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Re: If root{32x} = root(2x) +1, then 4x^2 =
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20 Mar 2016, 17:25
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), 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(32x)1 = sqrt(2x)  square everything 2x = 32x  2*sqrt(32x)+1 2x=4  2x  2*sqrt(32x) 44x=2*sqrt(32x)  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 = 6x1 E



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Re: If root{32x} = root(2x) +1, then 4x^2 =
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08 Mar 2018, 17:48
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), 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(32x)=sqrt(2) +1. Squaring on both sides: 32x=2x+1+2*sqrt(2x) Simplifying we get (4x)^2 = 6x1 Hence E.
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Re: If root{32x} = root(2x) +1, then 4x^2 =
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02 May 2018, 07:51
Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? thank you for your kind attention to my question



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If root{32x} = root(2x) +1, then 4x^2 =
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02 May 2018, 08:17
dave13 wrote: Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? 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{32x} = root(2x) +1, then 4x^2 =
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02 May 2018, 09:36
Bunuel wrote: dave13 wrote: Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? 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 \((\sqrt{32x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\) so we get \((32x)^2 = (2x1)^2\) now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\) we get \((32x) ( 32x) = (2x+1) (2x+1)\) 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{32x} = root(2x) +1, then 4x^2 =
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02 May 2018, 11:28
dave13 wrote: Bunuel wrote: dave13 wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 \(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\). Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? 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 \((\sqrt{32x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\) so we get \((32x)^2 = (2x1)^2\) now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\) we get \((32x) ( 32x) = (2x+1) (2x+1)\) 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 \((\sqrt{32x})^2 =32x\). 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{32x} = root(2x) +1, then 4x^2 =
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03 May 2018, 09:31
Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this \(1^2  2 * 1 *(2x) +2x^2 = 2x\) > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? what am i doing wrong , can you please explain in two or three words many thanks!



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Re: If root{32x} = root(2x) +1, then 4x^2 =
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03 May 2018, 10:16



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If root{32x} = root(2x) +1, then 4x^2 =
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03 May 2018, 11:19
Bunuel wrote: dave13 wrote: Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this 1^2  2 * 1 * (2x) + 2x^2 = 2x > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? 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{32x} = root(2x) +1, then 4x^2 =
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03 May 2018, 11:21
dave13 wrote: Bunuel wrote: dave13 wrote: Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this 1^2  2 * 1 * (2x) + 2x^2 = 2x > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? 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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Re: If root{32x} = root(2x) +1, then 4x^2 =
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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 \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E.




Re: If root{32x} = root(2x) +1, then 4x^2 = &nbs
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