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19 Nov 2023, 20:02
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E
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Difficulty:
65%
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Question Stats:
67% (02:12) correct
33%
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wrong
based on 1168
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If \(x - 10 = \sqrt{x} + \sqrt{10}\), what is the value of x?
(A) \(1 + √10\)
(B) \(10 + √10\)
(C) \(11 + √10\)
(D) \(10 + 2√10\)
(E) \(11 + 2√10\)
(A) \(1 + √10\)
(B) \(10 + √10\)
(C) \(11 + √10\)
(D) \(10 + 2√10\)
(E) \(11 + 2√10\)
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19 Nov 2023, 23:39
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It can be inferred from the prompt that this question wants to test use of: \(a^2-b^2=(a-b)(a+b)\)
\( x-10 = (\sqrt{x})^2 - (\sqrt{10})^2 = (\sqrt{x}-\sqrt{10})(\sqrt{x}+\sqrt{10})\)
Since the question says, \(x-10 = \sqrt{x}+\sqrt{10}\) ; cross them out in the above equation
\(\sqrt{x}-\sqrt{10}=1\)
or \(\sqrt{x}=1+\sqrt{10}\)
Then square this to get x : \(1+10+2\sqrt{10}\)
Answer E
\( x-10 = (\sqrt{x})^2 - (\sqrt{10})^2 = (\sqrt{x}-\sqrt{10})(\sqrt{x}+\sqrt{10})\)
Since the question says, \(x-10 = \sqrt{x}+\sqrt{10}\) ; cross them out in the above equation
\(\sqrt{x}-\sqrt{10}=1\)
or \(\sqrt{x}=1+\sqrt{10}\)
Then square this to get x : \(1+10+2\sqrt{10}\)
Answer E
19 Nov 2023, 21:57
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Arghi
\(x^2 - y^2 = (x+y)(x-y)\)
\(x - 10 = √x + √10\)
\((\sqrt{x})^2 - (\sqrt{10})^2 = \sqrt{x} + \sqrt{10}\)
\((\sqrt{x} + \sqrt{10})(\sqrt{x} - \sqrt{10}) = \sqrt{x} + \sqrt{10}\)
\((\sqrt{x} + \sqrt{10})(\sqrt{x} - \sqrt{10}) -(\sqrt{x} + \sqrt{10}) = 0\)
\((\sqrt{x} + \sqrt{10})[(\sqrt{x} - \sqrt{10}) - 1] = 0\)
As \((\sqrt{x} + \sqrt{10})\) is positive, \((\sqrt{x} - \sqrt{10}) - 1 = 0\)
\((\sqrt{x} - \sqrt{10}) - 1 = 0\)
\(\sqrt{x} = \sqrt{10}+1\)
Squaring on both sides
\(x = 10 + 1 + 2\sqrt{10}\)
\(x = 11 + 2\sqrt{10}\)
Option E
