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GMAT 1: 760 Q51 V42
Posts: 10,057
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:27) correct
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(02:58)
wrong
based on 328
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If \(x^2 + 4x + 1 = 0\), what is the value of \(x^2 + \frac{1}{x^2}\)?
A. 10
B. 12
C. 14
D. 16
E. 18
A. 10
B. 12
C. 14
D. 16
E. 18
MathRevolution
i.e. \(x^2 + 4x + 1 = 0\)
Dividing both sides by x we get
i.e. \(x + 4 + \frac{1}{x} = 0\)
i.e. \(x + \frac{1}{x} = -4\)
Squaring both sides we get
\(x^2 + \frac{1}{x^2}+2*x*\frac{1}{x} = 16\)
i.e. \(x^2 + \frac{1}{x^2} = 14\)
Answer: Option C
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General Discussion
GMAT 1: 760 Q51 V42
Posts: 10,057
=>
\(x^2 + 4x + 1 = 0\)
\(=> x + 4 + \frac{1}{x} = 0\)
\(=> x + \frac{1}{x} = -4\)
So, \((x + \frac{1}{x})^2 = (-4)^2\) and \(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16.\)
Therefore,
\(x^2 +2 + \frac{1}{x^2} = 16\)
and \(x^2 + \frac{1}{x^2} = 14.\)
Therefore, the answer is C.
Answer: C
\(x^2 + 4x + 1 = 0\)
\(=> x + 4 + \frac{1}{x} = 0\)
\(=> x + \frac{1}{x} = -4\)
So, \((x + \frac{1}{x})^2 = (-4)^2\) and \(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16.\)
Therefore,
\(x^2 +2 + \frac{1}{x^2} = 16\)
and \(x^2 + \frac{1}{x^2} = 14.\)
Therefore, the answer is C.
Answer: C
Posts: 1,324
X^2 + 4X + 1 = 0
X^2 + 4X + 4 - 4 + 1 = 0
(X + 2)^2 - 3 = 0
(X + 2)^2 = 3
Take the square root of each side ——-> taking the square root of an unknown binomial squared will result in the absolute value
[X + 2] = sqrt(3)
Case 1: X + 2 = sqrt(3)
X = sqrt(3) - 2
Case 2: -(X + 2) = sqrt(3)
X + 2 = (-) sqrt(3)
X = (-) sqrt(3) - 2
Since in the expression asked for we are squaring the given value of X, both roots will give the same value
(I)
(X)^2 = (sqrt(3) - 2)^2 = 7 - 4*sqrt(3) —-(I)
(II)
1 / (X)^2 = 1 / (7 - 4*sqrt(3))
We need to multiply the NUM and DEN by the conjugate of the DEN ———>
(1 / 7 - 4*sqrt(3))
*
(7 + 4*sqrt(3)) / (7 + 4*sqrt(3))
=
(7 + 4 * sqrt(3))
____________
(7 - 4*sqrt(3)) * (7 + 4*sqrt(3))
=
7 + 4*sqrt(3)
__________
49 - 16*3
=
7 + 4*sqrt(3) ——- (II)
And
(X)^2 + (1 / X^2) = (I) + (II) =
7 - 4*sqrt(3) + 7 + 4*sqrt(3) =
14
Posted from my mobile device
X^2 + 4X + 4 - 4 + 1 = 0
(X + 2)^2 - 3 = 0
(X + 2)^2 = 3
Take the square root of each side ——-> taking the square root of an unknown binomial squared will result in the absolute value
[X + 2] = sqrt(3)
Case 1: X + 2 = sqrt(3)
X = sqrt(3) - 2
Case 2: -(X + 2) = sqrt(3)
X + 2 = (-) sqrt(3)
X = (-) sqrt(3) - 2
Since in the expression asked for we are squaring the given value of X, both roots will give the same value
(I)
(X)^2 = (sqrt(3) - 2)^2 = 7 - 4*sqrt(3) —-(I)
(II)
1 / (X)^2 = 1 / (7 - 4*sqrt(3))
We need to multiply the NUM and DEN by the conjugate of the DEN ———>
(1 / 7 - 4*sqrt(3))
*
(7 + 4*sqrt(3)) / (7 + 4*sqrt(3))
=
(7 + 4 * sqrt(3))
____________
(7 - 4*sqrt(3)) * (7 + 4*sqrt(3))
=
7 + 4*sqrt(3)
__________
49 - 16*3
=
7 + 4*sqrt(3) ——- (II)
And
(X)^2 + (1 / X^2) = (I) + (II) =
7 - 4*sqrt(3) + 7 + 4*sqrt(3) =
14
Posted from my mobile device
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