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11 Jul 2025, 02:50
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(15 - 10 * (2)^0.5)^ 0.5
Let the two integers be a and b
=> (a)^0.5 - (b)^0.5 = (15 - 10 * (2)^0.5)^ 0.5
Squaring both sides we get,
((a)^0.5 - (b)^0.5)^2 = ((15 - 10 * (2)^0.5)^ 0.5)^2
=> a + b - 2((ab)^0.5) = 15 - 10*((2)^0.5)
Since we know that a and b are integers,
a + b = 15
-2 ((xy)^0.5) = - 10*((2)^0.5)
=> ab = 50
Solving a and b we get,
a = 10, b = 5
We need square of the difference between these two integers,
=> (a - b)^2 = (10-5)^2 = 5^2 = 25
C. 25
Let the two integers be a and b
=> (a)^0.5 - (b)^0.5 = (15 - 10 * (2)^0.5)^ 0.5
Squaring both sides we get,
((a)^0.5 - (b)^0.5)^2 = ((15 - 10 * (2)^0.5)^ 0.5)^2
=> a + b - 2((ab)^0.5) = 15 - 10*((2)^0.5)
Since we know that a and b are integers,
a + b = 15
-2 ((xy)^0.5) = - 10*((2)^0.5)
=> ab = 50
Solving a and b we get,
a = 10, b = 5
We need square of the difference between these two integers,
=> (a - b)^2 = (10-5)^2 = 5^2 = 25
C. 25
11 Jul 2025, 03:09
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Let the square roots of the two integers be \(\sqrt{a}\),\(\sqrt{b}\) and (\(\sqrt{a} >\sqrt{b}\))
Then,
\(\sqrt{a}\)-\(\sqrt{b}\)=\(\sqrt{15-10sqrt2}\)
Squaring on both sides:
\((\sqrt{a}-\sqrt{b})^2 = 15-10\sqrt{2}\)
==> \(a+b-2\sqrt{ab}=15-10\sqrt{2}\)
On comparing rational and irrational parts:
==> \(a+b= 15\),----eq1
\(ab= 50\)
We know ,\((a+b)^2= (a-b)^2 + 4ab\)
==> \(225= (a-b)^2 + 200\)
==> \((a-b)^2= 25\)
==> \((a-b) = +5 or -5\)----eq2
From eq1 and eq2 On solving,we get
==> a = 10, b = 5
The square of the difference between the integers=\((10-5)^2=5^2\)= \(25\)
Final answer Option C
Then,
\(\sqrt{a}\)-\(\sqrt{b}\)=\(\sqrt{15-10sqrt2}\)
Squaring on both sides:
\((\sqrt{a}-\sqrt{b})^2 = 15-10\sqrt{2}\)
==> \(a+b-2\sqrt{ab}=15-10\sqrt{2}\)
On comparing rational and irrational parts:
==> \(a+b= 15\),----eq1
\(ab= 50\)
We know ,\((a+b)^2= (a-b)^2 + 4ab\)
==> \(225= (a-b)^2 + 200\)
==> \((a-b)^2= 25\)
==> \((a-b) = +5 or -5\)----eq2
From eq1 and eq2 On solving,we get
==> a = 10, b = 5
The square of the difference between the integers=\((10-5)^2=5^2\)= \(25\)
Final answer Option C
11 Jul 2025, 03:19
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Given,
To find
(a - b)[sup]2[/sup] =?
Solution:
Squaring both sides,
a + b - 2√ab =
equating rational & irrational part, we have,
a + b = 15 & ab = 50
Using sum of roots & product of root to write the quadratic equation, we get,
x[sup]2 [/sup]- 15 x + 50 = 0
Solving, we get.
x = 10, 5
So, square of difference of two integer = ( 10 -5)[sup]2 [/sup]= 25
Ans: C
GMAT-Club-Forum-o7sjh2zq.png [ 1.06 KiB | Viewed 298 times ]
GMAT-Club-Forum-6p7vgjh9.png [ 1.06 KiB | Viewed 297 times ]
GMAT-Club-Forum-igud7tik.png [ 648 Bytes | Viewed 299 times ]
To find
(a - b)[sup]2[/sup] =?
Solution:
Squaring both sides,
a + b - 2√ab =
equating rational & irrational part, we have,
a + b = 15 & ab = 50
Using sum of roots & product of root to write the quadratic equation, we get,
x[sup]2 [/sup]- 15 x + 50 = 0
Solving, we get.
x = 10, 5
So, square of difference of two integer = ( 10 -5)[sup]2 [/sup]= 25
Ans: C
Bunuel
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GMAT-Club-Forum-o7sjh2zq.png [ 1.06 KiB | Viewed 298 times ]
Attachment:
GMAT-Club-Forum-6p7vgjh9.png [ 1.06 KiB | Viewed 297 times ]
Attachment:
GMAT-Club-Forum-igud7tik.png [ 648 Bytes | Viewed 299 times ]
