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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