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# Which of the following equations has 1 + √2 as one of its roots?

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Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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15 Jun 2016, 01:40
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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

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Re: Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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03 Sep 2016, 09:00
15
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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

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.

Since we know that 1 - √2 is a root of the quadratic equation, we can let:

r = 1 + √2

and

s = 1 - √2

Thus, r + s = (1 + √2) + (1 - √2) = 2 and rs = (1 + √2)(1 - √2) = 1 – 2 = -1.

Therefore, the quadratic equation must be x^2 – 2x – 1 = 0.

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Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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15 Jun 2016, 02:03
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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
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Re: Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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15 Jun 2016, 18:46
A and C are automatically out as they are roots of 1 and -1 respectively

for rest of option substitute 1 - root(2)...which ever equation gets satisfied is the answer.

Only equation D satisfies

Ans: d
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Re: Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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15 Jun 2016, 23:23
7
6
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 $$-2\sqrt{2}$$.

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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 06 Jun 2014 Posts: 90 Location: United States Concentration: Finance, General Management GMAT 1: 450 Q27 V21 GPA: 3.47 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 03 Jul 2016, 16:10 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. How can I solve option A, using Vieta's Theorem? _________________ 1) Kaplanprep 450 Q27 V21 2) Manhattan 530 Q35 V28 3) GmatPrep 450 Q33, V19 4) Veritas 460 Q31, V23 5) Veritas 440 Q 30, V21 6) Veritas 500 Q34, V 25 7) Gmat 420 Q27, V23 8) Veritas 520 Q36, V26 2/2 9) Veritas 540 Q37, V28 4/19 10)Manhattan 560 Q40, V28 4/28 Intern Joined: 04 Jun 2016 Posts: 7 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 03 Jul 2016, 20:58 1 I like to think about this question in this way: As irrational roots occur in pairs, the other root has to be 1-\sqrt{2}. So the unknown coefficients would be (-)sum of roots and (+)product of roots. This gives us D. Is this a good approach for these kind of questions? Intern Joined: 03 Jun 2015 Posts: 3 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 04 Jul 2016, 01:05 zxcvbnmas wrote: 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. How can I solve option A, using Vieta's Theorem? Viete theorem always works. But, if you don't have information about the solutions of the quadratic equation, the Viete theorem's use not always facilitates mental calculations. Option A is exactly the case. Intern Joined: 17 Jul 2016 Posts: 35 Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 31 Jul 2016, 10:11 if x^2=px+q then x=.5p+_root(.25p^2+q) This is very convenient to use in place of the quadratic formula when the coefficient of the linear term is even. a) x^2=-2x+1 ; x=-1+_root(1+1) b) x^2=2x-1; x=1+_root(1-1) c) x^2=-2x-1; x=-1+_root(1-1) d) x^2=2x+1; x=1+_root(1+1) The answer has to be d. Director Status: Preparing for GMAT Joined: 25 Nov 2015 Posts: 742 Location: India GPA: 3.64 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 26 Nov 2016, 05:57 1 1 Another method to solve this question is : x=$$(-b±\sqrt{(b^2-4ac)})/2a$$ , 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. _________________ Please give kudos, if you like my post When the going gets tough, the tough gets going... Manager Joined: 03 Jan 2017 Posts: 178 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 25 Mar 2017, 09:40 1 let's just calculate answer for each: 1) (-2+-8)/2 we see that we need negative b, so that the root could be (2+-8)/2 D is fine Manager Joined: 20 Apr 2014 Posts: 100 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 26 Mar 2017, 09:18 1 Hi Bunuel Could you please put your input here ? I can not get the concept or people are talking about something I already miss BR. Director Joined: 26 Aug 2016 Posts: 687 Location: India Concentration: Strategy, Marketing GMAT 1: 690 Q50 V33 GMAT 2: 700 Q50 V33 GMAT 3: 730 Q51 V38 GPA: 4 WE: Consulting (Consulting) Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 26 Mar 2017, 09:54 1 Use the formula roots = { -b +/- sqRt(b^2 - 4ac) } / 2a Math Expert Joined: 02 Sep 2009 Posts: 47983 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 28 Mar 2017, 02:59 1 5 hatemnag wrote: Hi Bunuel Could you please put your input here ? I can not get the concept or people are talking about something I already miss BR. Check the links below: Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm Theory on Algebra: http://gmatclub.com/forum/algebra-101576.html Algebra - Tips and hints: http://gmatclub.com/forum/algebra-tips- ... 75003.html DS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=29 PS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=50 Hope it helps. _________________ Manager Joined: 23 Jan 2016 Posts: 211 Location: India GPA: 3.2 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 09 Apr 2017, 07:34 1 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. how did you get 1 - √2 as the second root?? should it not be -(1+√2) or -1-√2?? Bunuel, I would really appreciate your help here, im trying to solve this question using vieta's theorum. Thank you. Intern Joined: 02 Oct 2013 Posts: 6 Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 13 May 2017, 05:56 use this equation a^2-b^2 = (a-b)(a+b) $$x^2-2x-1 = x^2-2x+1-2$$ = $$(x-1)^2-\sqrt{2}^2 = (x-1+\sqrt{2})(x-1-\sqrt{2})$$ thus, x = 1+\sqrt{2} and 1-\sqrt{2} Intern Joined: 28 Dec 2010 Posts: 23 Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 26 Jul 2017, 13:16 2 2 This problem involves irrational roots of a quadratic equation: For a quadratic equation: ax^2+bx+c = 0 where, the coefficients a,b and c are real. If we have one root as m +√n than it will also have a conjugate root m -√n where m, n are rational and n is not a perfect square. According to the question, one root is 1+√2 than the other root will be its conjugate i.e. 1-√2. Now, if we know the (sum of the roots) and (product of the roots) we can create a quadratic equation as: x^2 - (sum of roots) x + product of the roots = 0 Sum of roots = (1+√2) + (1-√2) = 2 Product of roots = (1+√2) * (1-√2) = 1 - (√2)^2 = -1 Quadratic Equation = x^2 -2x+1 = 0 i.e. option D. Manager Joined: 08 Apr 2017 Posts: 83 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 07 Nov 2017, 16:49 5 1 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 x = 1 + √2 x - 1 = √2 Squaring both sides (x - 1)^2 = (√2)^2 X^2 + 1 - 2x = 2 x^2 – 2x – 1 = 0 Option D EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12196 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink] ### Show Tags 10 Jan 2018, 20:46 3 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Which of the following equations has 1 + √2 as one of its roots?  [#permalink]

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02 May 2018, 22:59
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I just plugged the approximate value of 1+ √2 as 1+ approx 1.4= approx 2.4 and tested on answer choices.

D) satisfies as 5. something - 4.something - 1 is approx 0.
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Re: Which of the following equations has 1 + √2 as one of its roots? &nbs [#permalink] 02 May 2018, 22:59

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