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28 Feb 2024, 11:24
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If \(x > y > 0\), and \(x^2 + y^2 = 6xy\), what is the value of \(\frac{x - y}{x + y}\)?
A. \(2\sqrt{2}\)
B. \(\sqrt{2}\)
C. \(1\)
D. \(\frac{\sqrt{2}}{2}\)
E. \(\frac{1}{2}\)
A. \(2\sqrt{2}\)
B. \(\sqrt{2}\)
C. \(1\)
D. \(\frac{\sqrt{2}}{2}\)
E. \(\frac{1}{2}\)
28 Feb 2024, 11:24
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Official Solution:
If \(x > y > 0\), and \(x^2 + y^2 = 6xy\), what is the value of \(\frac{x - y}{x + y}\)?
A. \(2\sqrt{2}\)
B. \(\sqrt{2}\)
C. \(1\)
D. \(\frac{\sqrt{2}}{2}\)
E. \(\frac{1}{2}\)
Since we need to find the value of \(\frac{x - y}{x + y}\), let's try transforming \(x^2 + y^2 = 6xy\) such that we get both \(x - y\) and \(x + y\) from it. For that, we can rearrange it to obtain the square of a difference and the square of a sum.
Subtract \(2xy\) from both sides of \(x^2 + y^2 = 6xy\):
Now, add \(2xy\) to both sides of \(x^2 + y^2 = 6xy\):
Hence, \(\frac{x - y}{x + y} = \frac{2\sqrt{xy}}{2\sqrt{2xy}} = \frac{1}{\sqrt{2}}\). Rationalize the fraction by multiplying the numerator and denominator by \(\sqrt{2}\) to get \(\frac{\sqrt{2}}{\sqrt{2}*\sqrt{2}} = \frac{\sqrt{2}}{2}\).
Answer: D
If \(x > y > 0\), and \(x^2 + y^2 = 6xy\), what is the value of \(\frac{x - y}{x + y}\)?
A. \(2\sqrt{2}\)
B. \(\sqrt{2}\)
C. \(1\)
D. \(\frac{\sqrt{2}}{2}\)
E. \(\frac{1}{2}\)
Since we need to find the value of \(\frac{x - y}{x + y}\), let's try transforming \(x^2 + y^2 = 6xy\) such that we get both \(x - y\) and \(x + y\) from it. For that, we can rearrange it to obtain the square of a difference and the square of a sum.
Subtract \(2xy\) from both sides of \(x^2 + y^2 = 6xy\):
- \(x^2 + y^2 - 2xy= 6xy - 2xy\)
\((x-y)^2 = 4xy\)
\(x-y=2\sqrt{xy}\). Note here that both sides will be positive and defined since we are given that \(x > y > 0\).
Now, add \(2xy\) to both sides of \(x^2 + y^2 = 6xy\):
- \(x^2 + y^2 + 2xy= 6xy + 2xy\)
\((x+y)^2 = 8xy\)
\(x+y=2\sqrt{2xy}\). And again, both sides will be positive and defined since we are given that \(x > y > 0\).
Hence, \(\frac{x - y}{x + y} = \frac{2\sqrt{xy}}{2\sqrt{2xy}} = \frac{1}{\sqrt{2}}\). Rationalize the fraction by multiplying the numerator and denominator by \(\sqrt{2}\) to get \(\frac{\sqrt{2}}{\sqrt{2}*\sqrt{2}} = \frac{\sqrt{2}}{2}\).
Answer: D
24 May 2024, 01:39
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I think this is a high-quality question and I agree with explanation.
