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by AbeNeedsAnswers » Wed Jul 19, 2017 10:08 pm
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

D
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by Jay@ManhattanReview » Wed Jul 19, 2017 11:59 pm
AbeNeedsAnswers 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

D
We see that one of the roots is (1 + √2), which is not a whole number or it is an irrational root, thus the other root of the quadratic equation must be the conjugate of (1 + √2) = (1 - √2).

To get the conjugate, reverse the sign of the irrational part. Since in the given root, (1 + √2), the sign of the irrational part is +, so we must make it - to get its conjugate.

Thus, the two roots are: (1 + √2) and (1 - √2)

You must know that any quadratic equation can be written as:

x^2 - (Sum of the roots)*x + (Product of the roots) = 0

For the given roots,

Sum of the roots = (1 + √2) + (1 - √2) = 2, and
Product of the roots = (1 + √2) x (1 - √2) = 1^2 - (√2)^2 = 1 - 2 = -1

Thus, the quadratic equation is x^2 - (2)*x + (-1) = 0

[spoiler]x^2 - 2x -1 = 0[/spoiler]

The correct answer: D

Hope this helps!

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by Jay@ManhattanReview » Thu Jul 20, 2017 12:10 am
AbeNeedsAnswers 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

D
Another approach could be plug-in the value. Though handling the square of (1 + √2) would be difficult, doing a step at a time and analyzing at each level would save time and take us home.

Let's take option A.

A. x^2 + 2x - 1 = 0

(1 + √2)^2 + 2(1 + √2) - 1 = 0

(1 + 2 + 2√2) + (2 + 2√2) -1 = 0

To get '0' on the RHS, every term must cancel; however, we see that only '1' gets canceled, while '2' and '2√2' gets added. To cancel them, the sign of the middle term (shown in blue) of the quadratic equation x^2 + 2x - 1 = 0 must be negative.

Thus, the equations should be x^2 - 2x - 1 = 0.

Let's verify...

x^2 - 2x - 1 = 0

(1 + √2)^2 - 2(1 + √2) - 1 = 0

1 + 2 + 2√2 - 2 - 2√2 - 1 = 0

0 = 0.

Thus, the correct answer is Option D.

Hope this helps!

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by [email protected] » Thu Jul 20, 2017 11:32 am
Hi AbeNeedsAnswers,

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: D

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by GMATGuruNY » Thu Jul 20, 2017 11:49 am
AbeNeedsAnswers 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
A, B, C and D imply the following:
A) x² + 2x = 1
B) x² - 2x = -1
C) x² + 2x = -1
D) x² - 2x = 1.

Since these options all include x² and 2x , calculate x² = (1 + √2)² and 2x = 2(1 + √2):
x² = (1 + √2)² = 1² + √2² + 2(1)(√2) = 3 + 2√2.
2x = 2(1 + √2) = 2 + 2√2.

If we subtract the blue expressions from the red expressions, we get:
x² - 2x = (3 + 2√2) - (2 + 2√2)
x² - 2x = 1.

The correct answer is D.
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