InterviewSolution
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InterviewSolution
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Statement-1: `If a, b, c in Q and 2^(1//3)` is a root of `ax^(2) + bx + c = 0`, then a = b = c = 0. Statement-2: A polynomial equation with rational coefficients cannot have irrational roots.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C Multiplying both sides of `ax^(2) + bx + c = 0` by the LCM of denominators of rational numbers a, b and c, we obtain a quadratic equation with integer coefficients. So, let us assume that a, b, c are integers and they do not have any common factors. It is given that `2^(1//3)` satisfies `ax^(2) + bx + c = 0`. `therefore" "2^(2//3) a+2^(1//3)b+c = 0` `rArr" "c =-(2^(1//3)a+2^(1//3)b)` `rArr" "c^(3) =-{4a^(3) + 2b^(3)+6ab(2^(2//3) a+2^(1//3)b)}` `rArr" "c^(3) = - 4a^(3) - 2b^(3) + 6abc` `rArr" "c^(3) =-2(2a^(3) + b^(3) - 3abc)` `rArr" "2|c^(3)` `rArr" "2|c` `rArr" "c = 2 c_(1)` for some integer `c_(1)`. Puttion `c = 2c_(1)` in (i), we get `8c_(1)^(3) = -4a^(3) - 2b^(3) + 6 abc_(1)` `rArr" "4c_(1)^(3) = -(2a^(3)+b^(3) - 6 abc_(1))` `rArr" "b^(3) = -2(a^(3) + 2c_(1)^(3) - 3 abc_(1))` `rArr" "2|b^(3) rArr 2| b rArr b = 2b_(1)` for some integer `b_(1)`. Putting `b = 2b_(1) and c = 2c_(1)` in (i), we get `8c_(1)^(3) = - 4a^(3) - 16 b_(1)^(3) + 24 ab_(1) c_(1)` `rArr" "a^(3) = - 2(c_(1)^(3)+2b_(1)^(3) - 3ab_(1) c_(1)) rArr 2|a^(3) rArr 2|a` This is a contradiction. Therefore, a = b = c = 0. So, Statement-1 is true. Statement-2 is false, because `x^(3) - 2 = 0` is a polynomial equation with rational coefficients having `2^(1//3)` as its root. |
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