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

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

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

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

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

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

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

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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.

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

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

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

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

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Use the formula roots = { -b +/- sqRt(b^2 - 4ac) } / 2a
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Re: Which of the following equations has 1 + √2 as one of its roots? [#permalink]

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

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

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

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

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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


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

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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....

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

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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?   [#permalink] 02 May 2018, 22:59

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