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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Which of the following equations is satisfied by x=1+âˆš2? tagged by: Max@Math Revolution ##### This topic has 3 expert replies and 1 member reply ### GMAT/MBA Expert ## Which of the following equations is satisfied by x=1+âˆš2? ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question] Which of the following equations is satisfied by x=1+âˆš2? 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 - 2 = 0 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

Which of the following equations is satisfied by x=1+âˆš2?

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 - 2 = 0
Since the options are quadratic equations and x = 1 + âˆš2 is a linear equation, we must find a way to convert into a quadratic equation such that we have a term with x.

So, we have x = 1 + âˆš2

=> x - 1 = âˆš2

Squaring both the sides, we get

(x - 1)^2 = (âˆš2)^2

x^2 - 2x + 1 = 2

x^2 - 2x - 1 = 0

Hope this helps!

-Jay
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### GMAT/MBA Expert

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

Which of the following equations is satisfied by x=1+âˆš2?

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 - 2 = 0
Alternate approach:

For any quadratic in the form xÂ² + bx + c = 0, where b and c are rational numbers:
If one of the roots is m+âˆšn, then the other root is m-âˆšn.
Sum of the roots = -b
Product of the roots = c

Here, since one of the roots is 1+âˆš2, the other root is 1-âˆš2.
Thus:

-b = sum of the roots
-b = (1+âˆš2) + (1-âˆš2)
-b = 2
b = -2

c = product of the roots
c = (1+âˆš2)(1-âˆš2)
c = 1-2
c = -1

xÂ² - 2x -1 = 0

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### Top Member

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Looking at option B and D
$$B=x^2-2x+1:\ its\ roots\ 1\ and\ 1$$
because $$\left(x-1\right)^2=x^2-2x+1=0\$$
$$D=x^2+2x+1=0;\ its\ root\ are\ -1\ and\ -1$$
because $$\left(x+1\right)^2=x^2+2x+1=0$$
If we substitute
$$1+\sqrt{2}\ into\ these\ equation,\ it\ is\ne0$$
The remaining 3 option have
$$x^2\ where\ \ x=1+\sqrt{2}=\left(1+\sqrt{2}\right)^2=2\sqrt{2}$$
This means that for
$$1+\sqrt{2}$$ to satisfy any of these equation must be =0 thus, $$x^2$$ which is $$2\sqrt{2}$$ to get 0
Option A is the only equation that satisfy these criteria.
$$x=1+\sqrt{2}$$
$$x-1=\sqrt{2}$$
$$x^2-2x-1=0$$
$$2\sqrt{2}-2\left(x-1\right)=0$$
$$2\sqrt{2}-2\sqrt{2}=0$$
$$OR\ \ \ x=1+\sqrt{2}\ \ \$$
$$x-1=\sqrt{2}$$
$$square\ both\ sides\ \ \left(x-1\right)^2=\left(\sqrt{2}\right)^2$$
$$\left(x-1\right)\left(x-1\right)=2$$
$$x^2-x-x+1=2$$
$$x^2-2x+1=2$$
$$x^2-2x+1-2=0$$
$$x^2-2x-1=0\ \$$
$$answer\ is\ Option\ A$$

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=>
x=1+âˆš2
=> x - 1 = âˆš2
=> (x-1)^2 = (âˆš2)^2
=> x^2 - 2x + 1 = 2
=> x^2 - 2x - 1 = 0.

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