Maths-
General
Easy

Question

a = square root of 11+square root of 3, b = square root of 12+square root of 2  and c = square root of 6+square root of 4  , then which of the following holds true?
i) c > a > b     ii) a > b > c    iii) a > c > b     iv) b > a > c 

hintHint:

Hint:
If two numbers x and y are given and it is also given that x > y, then we say that x2 > y2 and vice versa is also possible.

The correct answer is: a > b > c


    Step 1 of 2:
                  a = square root of 11+square root of 3
    Squaring both sides
    a squared equals 11 plus 3 plus 2 square root of 33
    a squared equals 14 plus 2 square root of 33   .......  (1)
    b equals square root of 12 plus square root of 2
    Squaring both sides
    b squared equals 12 plus 2 plus 2 square root of 24
    b squared equals 14 plus 2 square root of 24 ..... (2)
    c equals square root of 6 plus square root of 4
    Squaring both sides
    c squared equals 6 plus 4 plus 2 square root of 24
    c squared equals 10 plus 2 square root of 24
    Step 2 of 2:
    We can see that from equation (1) and (2), a2 > b2 which means a > b, and from equation (2) and (3) b2 > c2 which means b > c.
    Final Answer:
    We found that a > b and b > c. So, we can conclude that a > b > c. Hence, option (ii) is correct.

    Note: 
    This question can simply be solved by finding the values of all the square roots and then using them in finding the values of a, b and c and then their values can be easily compared.

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