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Which of the following equations is satisfied by x=1+√2?

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by Max@Math Revolution » Thu Dec 06, 2018 11:47 pm

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[Math Revolution GMAT math practice question]

Which of the following equations is satisfied by x=1+√2?

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
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by Jay@ManhattanReview » Fri Dec 07, 2018 12:34 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Which of the following equations is satisfied by x=1+√2?

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 + √2 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 + √2

=> x - 1 = √2

Squaring both the sides, we get

(x - 1)^2 = (√2)^2

x^2 - 2x + 1 = 2

x^2 - 2x - 1 = 0

The correct answer: A

Hope this helps!

-Jay
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roots

by GMATGuruNY » Fri Dec 07, 2018 5:09 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Which of the following equations is satisfied by x=1+√2?

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 x² + bx + c = 0, where b and c are rational numbers:
If one of the roots is m+√n, then the other root is m-√n.
Sum of the roots = -b
Product of the roots = c

Here, since one of the roots is 1+√2, the other root is 1-√2.
Thus:

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

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

Resulting quadratic:
x² - 2x -1 = 0

The correct answer is A.
Last edited by GMATGuruNY on Tue Dec 11, 2018 10:06 am, edited 1 time in total.
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by deloitte247 » Sat Dec 08, 2018 1:08 am
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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by Max@Math Revolution » Sun Dec 09, 2018 5:49 pm
=>
x=1+√2
=> x - 1 = √2
=> (x-1)^2 = (√2)^2
=> x^2 - 2x + 1 = 2
=> x^2 - 2x - 1 = 0.

Therefore, the answer is A.
Answer: A
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