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10 Jul 2025, 18:12
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We are given :
[square_root(x)]-[square_root(y)]=[square_root(15-10[square_root]2)
squaring both side we will get
x+y -2* sq root (xy) =15-10 sq root(2)
we can re arrange and write it as
x+y -2 * sq root (xy)= 15-2 * sq root ( 50)
x+y=15
xy=50
we need to find (x-y)^2
which is (x+y)^2-4xy
(15)^2-4(50)
225-20=25 (option C)
[square_root(x)]-[square_root(y)]=[square_root(15-10[square_root]2)
squaring both side we will get
x+y -2* sq root (xy) =15-10 sq root(2)
we can re arrange and write it as
x+y -2 * sq root (xy)= 15-2 * sq root ( 50)
x+y=15
xy=50
we need to find (x-y)^2
which is (x+y)^2-4xy
(15)^2-4(50)
225-20=25 (option C)
Bunuel
lvillalon
Joined: 29 Jun 2025
Last visit: 25 Aug 2025
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Concentration: Operations, Entrepreneurship
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
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WE:Consulting (Consulting)
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
Posts: 88
10 Jul 2025, 18:20
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C. 25
Let a>b
sqrt(a)-sqrt(b) = sqrt(15 - 10sqrt(2))
(a-b)^2 = ?
(a-b)^2 = a^2 - 2ab + b^2
Resolution:
Square the first equation
a-2sqrt(ab)+b = 15 - 10sqrt(2)
a+b=15
2sqrt(ab)=10sqrt(2)
sqrt(ab)=5sqrt(2)
ab=50
(15-b)*b = 50
b^2-15b+50 = 0
(b-5)(b-10) = 0
b = 5 or 10
Depending on that, you've got
a = 10 or 5
Given a>b, a=10, b=5
(a-b)^2 = 25
Let a>b
sqrt(a)-sqrt(b) = sqrt(15 - 10sqrt(2))
(a-b)^2 = ?
(a-b)^2 = a^2 - 2ab + b^2
Resolution:
Square the first equation
a-2sqrt(ab)+b = 15 - 10sqrt(2)
a+b=15
2sqrt(ab)=10sqrt(2)
sqrt(ab)=5sqrt(2)
ab=50
(15-b)*b = 50
b^2-15b+50 = 0
(b-5)(b-10) = 0
b = 5 or 10
Depending on that, you've got
a = 10 or 5
Given a>b, a=10, b=5
(a-b)^2 = 25
Bunuel
10 Jul 2025, 19:34
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Let the 2 integers be A = \(x^2\), and B =\( y^2\),
then square roots of A and B are x, and y
To find :\( (A-B)^2 \)
Given : positive difference of the square roots of two integers = \(\sqrt{15 - 10\sqrt{2}}\)
\(|x-y| = \sqrt{15 - 10\sqrt{2}}\)
squaring both sides,
\(\\
(x-y)^2 = 15 - 10 \sqrt{2} \\
x^2 + y^2 - 2xy = 15 - 10\sqrt{2}\\
\)
we need to find x and y such that \(x^2 + y^2 = 15\), and \( 2xy = 10\sqrt{2} >> xy = 5\sqrt{2} \)
the sum of 15 can be split into 10 + 5, multiply\( \sqrt{10} * \sqrt{5} = \sqrt{2*5*5} = 5\sqrt{2} \)
so if we have \((\sqrt{10} - \sqrt{5} )^2 = 10 + 5 - 2*5 \sqrt{2} = 15 - 10\sqrt{2}\)
\(x=\sqrt{10}, y = \sqrt{5}\) or \(x =\sqrt{ 5}, y = \sqrt{10} \)
From this, A = 10, B = 5, or A = 5 and B = 10 >> ultimately we want to find (A-B)^2 so the difference will be the same - doesnt matter whether A is 10 or B is 10
Now we can find
\(\\
(A-B)^2 = (10-5)^2 = 5^2 = 25\\
\\
\)
Answer is C. 25
then square roots of A and B are x, and y
To find :\( (A-B)^2 \)
Given : positive difference of the square roots of two integers = \(\sqrt{15 - 10\sqrt{2}}\)
\(|x-y| = \sqrt{15 - 10\sqrt{2}}\)
squaring both sides,
\(\\
(x-y)^2 = 15 - 10 \sqrt{2} \\
x^2 + y^2 - 2xy = 15 - 10\sqrt{2}\\
\)
we need to find x and y such that \(x^2 + y^2 = 15\), and \( 2xy = 10\sqrt{2} >> xy = 5\sqrt{2} \)
the sum of 15 can be split into 10 + 5, multiply\( \sqrt{10} * \sqrt{5} = \sqrt{2*5*5} = 5\sqrt{2} \)
so if we have \((\sqrt{10} - \sqrt{5} )^2 = 10 + 5 - 2*5 \sqrt{2} = 15 - 10\sqrt{2}\)
\(x=\sqrt{10}, y = \sqrt{5}\) or \(x =\sqrt{ 5}, y = \sqrt{10} \)
From this, A = 10, B = 5, or A = 5 and B = 10 >> ultimately we want to find (A-B)^2 so the difference will be the same - doesnt matter whether A is 10 or B is 10
Now we can find
\(\\
(A-B)^2 = (10-5)^2 = 5^2 = 25\\
\\
\)
Answer is C. 25
