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Re: Which of the following equations has 1 + √2 as one of its roots?
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03 May 2018, 16:56
I found estimation to work the best. Sqrt2= 1.4 1+sqrt2 = 2.4 One root = 2.4 or x =2.4 X^2 = 5.76 2x= 4.8
Now you have all the values to solve. You are trying to see which expression will give you the closest value to zero.
C can be immediately crossed off because you are adding throughout the expression
When you subtract 5.76  4.8, you get a value close to 1 being left over.
Equation D matches what you are looking for.



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Re: Which of the following equations has 1 + √2 as one of its roots?
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22 Aug 2018, 01:48
fla wrote: Bunuel wrote: Which of the following equations has 1 + √2 as one of its roots?
A) x^2 + 2x – 1 = 0 B) x^2 – 2x + 1 = 0 C) x^2 + 2x + 1 = 0 D) x^2 – 2x – 1 = 0 E) x^2 – x – 1= 0 If x1 = 1 + √2 then x2 = 1  √2. By the Viete theorem: b = (x1 + x2) = (1 + √2 + 1  √2) = 2 c = x1 * x2 = (1 + √2)(1  √2) = 1  2 = 1 is the answer. This is one of the simplest methods. Thanks.
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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Mar 2019, 11:50
How can we use the sum of roots formula here? We don't have the value of a to be certain?



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Re: Which of the following equations has 1 + √2 as one of its roots?
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03 Jun 2019, 07:18
Looking through the answer responses, I have not heard of Viete theorem. I have also never heard of the rules mentioned by JeffTargetTestPrep (see below): To solve this problem, we need to use the following two facts: 1) If a quadratic equation has integer coefficients only, and if one of the roots is a + √b (where a and b are integers), then a  √b is also a root of the equation. 2) If r and s are roots of a quadratic equation, then the equation is of the form x^2 – (r +s)x + rs = 0. But thank you for making light of these rules and theorem. What I did and worked well for me, I used 1 + √2 and substitute into the answer choices. I started with answer choice C and made my way down and finally got the answer D this can be done relatively quickly! The idea is to ensure the left hand side (LHS) of the equation equals to the right hand side (RHS) of the equation, as follows: From answer choice D, x^2 – 2x – 1 = 0 x^2 – 2x – 1 = 0 x^2 2x = 1 (1 + √2)^2  2(1 + √2 ) = 1 (1+2+2√2)  (2 +2√2) =1 1+2+2√2  2 2√2 =1 From here you can cancel +2√2 and 2√2 1+22 = 1 1 = 1 LHS = RHS Answer D.



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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Jun 2019, 13:00
fla wrote: Bunuel wrote: Which of the following equations has 1 + √2 as one of its roots?
A) x^2 + 2x – 1 = 0 B) x^2 – 2x + 1 = 0 C) x^2 + 2x + 1 = 0 D) x^2 – 2x – 1 = 0 E) x^2 – x – 1= 0 If x1 = 1 + √2 then x2 = 1  √2.By the Viete theorem: b = (x1 + x2) = (1 + √2 + 1  √2) = 2 c = x1 * x2 = (1 + √2)(1  √2) = 1  2 = 1 is the answer. Hi Bunuel May I get your attention? But who knows x2 = 1  √2.?
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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Jun 2019, 13:23
VeritasKarishma wrote: Bunuel wrote: Which of the following equations has 1 + √2 as one of its roots?
A) x^2 + 2x – 1 = 0 B) x^2 – 2x + 1 = 0 C) x^2 + 2x + 1 = 0 D) x^2 – 2x – 1 = 0 E) x^2 – x – 1= 0 Note that (B) is (x  1)^2 so its roots are 1 and 1. (C) is (x + 1)^2 so its roots are 1 and 1. Now note that all 3 of the remaining options have x^2. When you put \(x = (1 + \sqrt{2})^2\) in them, you will get \((+2\sqrt{2})\) term. It should get canceled out by another term to get 0. So you should get \((2\sqrt{2})\) term. Option (D) has 2x which will give a term \(2\sqrt{2}\). So answer (D) Hi VeritasKarishmaIn the highlighted part, is the squared (^2) mistakenly put? Quote: (B) is (x  1)^2 so its roots are 1 and 1. (C) is (x + 1)^2 so its roots are 1 and 1. Could you explain the quoted part?
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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Jun 2019, 13:33
souvonik2k wrote: Another method to solve this question is : x=\((b±\sqrt{(b^24ac)})/2a\) , using equation for finding roots of quadratic equation where b is coefficient of x, a is coef of x^2 and c is constant. Substituting values for option A gives x=\(1±\sqrt{2}\) Since we need root as \(1+\sqrt{2}\), b must be negative, with other coef same as A which is option D Answer D. souvonik2kCould you explain a bit this one, please? thanks
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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Jun 2019, 13:43
EMPOWERgmatRichC wrote: Hi All, While this prompt might look a bit 'scary', you can answer it without doing a lot of complex math (but you need to pay attention to what each of the 5 equations implies (and whether you can actually get a sum of 0 in the end or not). To start, we're told that (1 + √2) is a 'root' of one of those equations, which means that when you plug that value in for X and complete the calculation, you will get 0 as a result.We know that √2 is greater than 1, so (1 + √2) will be GREATER than 2 (it's actually a little greater than 2.4, but you don't have to know that to answer this question). So, when you plug that value into X^2 (which appears in all 5 answers), you get a value that is GREATER than 4. To get that "greater than 4" value down to 0, we have to subtract something.... Also keep in mind that... (squaring a value greater than 2) > (doubling that same value) So subtracting 2X from X^2 would NOT be enough to get us down to 0... we would ALSO need to subtract the 1.... With those ideas in mind, there's only one answer that matches.... Final Answer: GMAT assassins aren't born, they're made, Rich EMPOWERgmatRichCThanks for the extraordinary explanation. But how do someone convinced that those equations are perfectly fine ( i mean they are not randomly written)?
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Re: Which of the following equations has 1 + √2 as one of its roots?
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17 Jun 2019, 14:36
Hi Asad, To start, NOTHING about an Official GMAT question is ever 'random'; every aspect of the question (including what appears in the 5 answer choices) was specifically chosen for a reason  and sometimes the reason is to present a pattern that you can take advantage of. From your question, I assume that you're asking about how we know for sure that the each of those equations actually does have a solution (or multiple solutions). Since we're looking for just one thing  the equation that has (1 + √2) as a root  it doesn't really matter what the other 4 equations are (since they won't have that root, none of them will be the answer to the question that was asked). GMAT assassins aren't born, they're made, Rich
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Re: Which of the following equations has 1 + √2 as one of its roots?
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19 Jun 2019, 03:32
Asad wrote: VeritasKarishma wrote: Bunuel wrote: Which of the following equations has 1 + √2 as one of its roots?
A) x^2 + 2x – 1 = 0 B) x^2 – 2x + 1 = 0 C) x^2 + 2x + 1 = 0 D) x^2 – 2x – 1 = 0 E) x^2 – x – 1= 0 Note that (B) is (x  1)^2 so its roots are 1 and 1. (C) is (x + 1)^2 so its roots are 1 and 1. Now note that all 3 of the remaining options have x^2. When you put \(x = (1 + \sqrt{2})^2\) in them, you will get \((+2\sqrt{2})\) term. It should get canceled out by another term to get 0. So you should get \((2\sqrt{2})\) term. Option (D) has 2x which will give a term \(2\sqrt{2}\). So answer (D) Hi VeritasKarishmaIn the highlighted part, is the squared (^2) mistakenly put? Quote: (B) is (x  1)^2 so its roots are 1 and 1. (C) is (x + 1)^2 so its roots are 1 and 1. Could you explain the quoted part? Yes, it should be When you put \(x = (1 + \sqrt{2})\) in them, you will get \((+2\sqrt{2})\) term. Basically, it means \(x^2 = ( 1 + \sqrt{2})^2 = 1 + 2 + 2*\sqrt{2}\) Recall that \((a + b)^2 = a^2 + b^2 + 2ab\) \((a  b)^2 = a^2 + b^2  2ab\) B) x^2 – 2x + 1 = 0 Look at the left side. It is (x  1)^2 because when you expand it, you get x^2  2x + 1. \((x  1)^2 = (x  1)*(x  1) = 0\) So x = 1, 1 C) x^2 + 2x + 1 = 0 Similarly, (x +1)^2 = x^2 + 2x + 1 So here (x + 1)(x + 1) = 0 x = 1, 1
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Re: Which of the following equations has 1 + √2 as one of its roots?
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