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10 Jul 2025, 09:50
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\(\sqrt{x}-\sqrt{y}=\sqrt{15-10(2)^(0.5)}\)
Squaring both sides:
=> \( x + y - 2\sqrt{x*y} = 15 - 10\sqrt{2}\)
=> \( x + y - 2\sqrt{x*y} = 15 - 2\sqrt{50}\)
=> x = 10, y=5
So \((x-y)^2=5^2=25\)
Hence answer is C
Squaring both sides:
=> \( x + y - 2\sqrt{x*y} = 15 - 10\sqrt{2}\)
=> \( x + y - 2\sqrt{x*y} = 15 - 2\sqrt{50}\)
=> x = 10, y=5
So \((x-y)^2=5^2=25\)
Hence answer is C
Bunuel
10 Jul 2025, 10:01
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Let m and n be the two integers.
We are told that,
ABS(\(\sqrt{m} - \sqrt{n}\)) = Sqrt(\(15 - 10 \sqrt{2}\))
Squaring both sides,
We get,
\((\sqrt{m} - \sqrt{n})^2\) = \(15 - 10 \sqrt{2}\)
\(m + n - 2 \sqrt{mn}\) = \(15 - 10 \sqrt{2}\)
Now we can compare the rational part of LHS with RHS, and same for the irrational part. This is going to give us a quadratic. However, to save time, let's just plug some numbers:
So we want values of m and n, such that m+n = 15 and \(2 \sqrt{mn}\) is \(10 \sqrt{2}\)
When m = 10 and n = 5, these criteria are satisfied.
Now we need to find the value of \((m-n)^2\),
Using m =10 and n=5, we will have this expression's value as 25.
Answer C.
We are told that,
ABS(\(\sqrt{m} - \sqrt{n}\)) = Sqrt(\(15 - 10 \sqrt{2}\))
Squaring both sides,
We get,
\((\sqrt{m} - \sqrt{n})^2\) = \(15 - 10 \sqrt{2}\)
\(m + n - 2 \sqrt{mn}\) = \(15 - 10 \sqrt{2}\)
Now we can compare the rational part of LHS with RHS, and same for the irrational part. This is going to give us a quadratic. However, to save time, let's just plug some numbers:
So we want values of m and n, such that m+n = 15 and \(2 \sqrt{mn}\) is \(10 \sqrt{2}\)
When m = 10 and n = 5, these criteria are satisfied.
Now we need to find the value of \((m-n)^2\),
Using m =10 and n=5, we will have this expression's value as 25.
Answer C.
10 Jul 2025, 10:18
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\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{15 - 10\sqrt{2}}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{10+5 - 2*5\sqrt{2}}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{10+5 - 2\sqrt{5*10}}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{(\sqrt{10}-\sqrt{5})^2}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(|\sqrt{10}-\sqrt{5}|\)
a=10, b=5
(a-b)^2=(10-5)^2=25
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{10+5 - 2*5\sqrt{2}}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{10+5 - 2\sqrt{5*10}}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(\sqrt{(\sqrt{10}-\sqrt{5})^2}\)
\(|\sqrt{a}-\sqrt{b}|\) = \(|\sqrt{10}-\sqrt{5}|\)
a=10, b=5
(a-b)^2=(10-5)^2=25
Bunuel
