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Of x= 2+root 3 and xy = 1 then x/root 2 + root x + y/root 2 - root y?
Verified Answer
Of x= 2+root 3 and xy = 1 then x/root 2 + root x + y/root 2 - root y?
Find the value of square roots of x and y. Then use rationalization of denominator to find the value of y.

x = 2 + √3  
xy = 1

y = 1 /(2 + √3) = (2 - √3)/[(2 - √3)(2 + √3) 
= 2 - √3

Let   √x = √a + √b
then   x = 2 + √3 = a + b + 2 √ab
so   a + b = 2    and   ab = 3 /4
(a - b)^2
= 4 - 3
= 1
= a - b = 1           
=>  a = 3/2     
=>  b = 1/2
so √x = (√3 +1)/√2
√y = 1/√x = (√3 - 1)/√2

Now substitute in the given expression:

This question is part of UPSC exam. View all Class 9 courses
Most Upvoted Answer
Of x= 2+root 3 and xy = 1 then x/root 2 + root x + y/root 2 - root y?
Given:
x = 2√3
xy = 1

To find:
x/√2 - √x * y/√2 - √y

Solution:

Step 1: Find the value of y
Given that xy = 1, we can solve for y:
y = 1/x

Step 2: Substitute the value of y in the expression
x/√2 - √x * y/√2 - √y

Substituting y = 1/x:
x/√2 - √x * (1/x)/√2 - √y

Step 3: Simplify the expression
x/√2 - √x/x√2 - √y
= x/√2 - √(x)/x√2 - √(1/x)

Step 4: Rationalize the denominator
To rationalize the denominator, we multiply the expression by its conjugate. In this case, the conjugate of x√2 is x√2.

(x/√2 - √(x)/x√2 - √(1/x)) * (x√2/x√2)
= (x * x√2)/(√2 * x√2) - (√(x) * x√2)/(x√2 * x√2) - (√(1/x) * x√2)/(x√2 * x√2)

Step 5: Simplify further
= (x^2√2)/(2x) - (√x * x√2)/(x^2) - (√(1/x) * x√2)/(x^2)

Step 6: Cancel out common factors
= √2/2 - √x/x - √(1/x)/x^2

Step 7: Substitute the value of x = 2√3
= √2/2 - √(2√3)/(2√3) - √(1/(2√3))/(2√3)
= √2/2 - √(2√3)/(2√3) - √(1/(2√3))/(2√3)

Step 8: Simplify the expressions
= √2/2 - (2√2√3)/(2√3) - (1/(2√2√3))/(2√3)

Step 9: Cancel out common factors
= √2/2 - √2/2 - 1/(4√2√3)

Step 10: Simplify further
= 0 - 1/(4√2√3)
= -1/(4√2√3)

Therefore, the simplified expression is -1/(4√2√3).
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Of x= 2+root 3 and xy = 1 then x/root 2 + root x + y/root 2 - root y?
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