If x+y=4, x²+y²=14 and x>y, Then the correct value of x and y is:

  1. 3, 1
  2. 2-√2, √3
  3. 2+√3, 2-√3
  4. 2+√3, 2√2
Anurag Mishra Professor Asked on 10th December 2015 in Maths.
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1 Answer(s)

Answer:  (3) 2 + √3, 2 – √3

Explanation:-
X + Y = 4 ……………………. (1)
Square on both side,
(X + Y)² = 4²
X² + Y² + 2 X Y = 16
Put the value of X² + Y² = 14,
Then, 14 + 2 X Y = 16
2 X Y = 16 – 14
2 X Y = 2
X Y = 1

(X – Y)² = X² + Y² – 2 X Y
Put the value of X Y = 1 & X² + Y² = 14,
Then, (X – Y)² = 14 – 2
(X – Y)² = 12
Square root on both side,
X – Y = 2√3 …………………..(2)

Add equation (1) & (2),
X + Y + X – Y = 4 + 2√3
2 X = 2 (2 + √3)
X = 2 + √3

Now, put the value of X = 2 + √3 in equation (1),
Then, 2 + √3 + Y = 4
Y = 4 – 2 – √3
Y = 2 – √3

Then, the value of X = X = 2 + √3 & Y = 2 – √3

Hence, the correct answer is option (3) 2 + √3, 2 – √3.

Anurag Mishra Professor Answered on 10th December 2015.
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