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04 Sep 2022, 10:35
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B
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Difficulty:
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Question Stats:
48% (02:35) correct
52%
(02:39)
wrong
based on 225
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What is the value of \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) ?
A. -4
B. -2
C. -1
D. 1
E. 2
A. -4
B. -2
C. -1
D. 1
E. 2
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21 Feb 2023, 06:44
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Bunuel
Official Solution:
What is the value of \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) ?
A. \(-4\)
B. \(-2\)
C. \(-1\)
D. \(1\)
E. \(2\)
We want to find the value of the expression \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\). To simplify this expression, we can start by squaring it.
- \(=((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10}))^2=\)
\(=(4-\sqrt{15})(4 + \sqrt{15})^2(\sqrt{6} - \sqrt{10})^2=\)
\(=(4-\sqrt{15})(4 + \sqrt{15})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Applying the difference of squares identity \((a - b)(a+b)=a^2-b^2\) to \((4-\sqrt{15})(4 + \sqrt{15})\) we can simplify to:
- \(=(16 -15)(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Now using the identity \((a - b)^2=a^2-2ab+b^2\) to simplify the squared term::
- \(=(4 + \sqrt{15})(6 - 2\sqrt{60}+10)=\)
\(=(4 + \sqrt{15})(16 - 4\sqrt{15})=\)
\(=(4 + \sqrt{15})4(4 - \sqrt{15})=\)
\(=4(16-15)=4\).
Since the square of the expression is 4, the expression itself must be either 2 or -2. However, since the product of two positive terms and one negative term is negative (\((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})=positive*positive*negative=negative\)), we know that \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) is negative. Therefore, the final answer is -2.
Answer: B
General Discussion
05 Sep 2022, 00:54
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Bunuel
Solution:
Let us assume \(a=(\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\)
Now, we need the value of a
If you notice, we have a square root over \(4-\sqrt{15}\) which if removed can be helpful. So, let us square both sides
\(⇒a^2=(\sqrt{4-\sqrt{15}})^2\times (4 + \sqrt{15})^2\times (\sqrt{6} - \sqrt{10})^2\)
\(⇒a^2=(4-\sqrt{15})\times(4+\sqrt{15})\times(4+\sqrt{15})\times(\sqrt{6}-\sqrt{10})^2\)
\(⇒a^2=(4^2-(\sqrt{15})^2)\times (4+\sqrt{15})\times (6+10-2\sqrt{60})\) (couple of identities \((a+b)(a-b)=a^2-b^2\) and \((a-b)^2=a^2+b^2-2ab)\)
\(⇒a^2=(16-15)\times (4+\sqrt{15})\times (16-2\sqrt{60})\)
\(⇒a^2=(4+\sqrt{15})\times (16-4\sqrt{15})\)
\(⇒a^2=(4+\sqrt{15})\times 4(4-\sqrt{15})\)
\(⇒a^2=4\times (4+\sqrt{15})(4-\sqrt{15})\)
\(⇒a^2=4(4^2-(\sqrt{15})^2)\)
\(⇒a^2=4\)
\(⇒a=-2\)
Note: we are ignoring \(a=2\) possibility because if you observe \(a=(\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) and the term \((\sqrt{6} - \sqrt{10})\) is negative and other two terms \((\sqrt{4-\sqrt{15}})\) and \((4 + \sqrt{15})\) are positive, so the ultimate result will be negative
Hence the right answer is Option B
