If all coefficents of a quadratic equation are rational, then the all irrational roots appear in "conjugate pairs". Your statement, "So the coefficients of a quadratic should be rational for the roots to be (irrational) pair", however, is not necessarily true. An obvious example is which has irrational coefficients but "irrational pair" roots, .
Suppose j is any rational number, in this case 3. I am a math beginner, but since j + any rational number is equal to a rational number [j + rational number = rational number], 3 + any rational number = a rational number. I know obvious. Now comes the challenge, is 3 + an irrational number equal to an irrational number? When we found out that 3 + any rational number = a rational number, so 3 + any irrational number can't be a rational number or j + irrational number does not equal a rational number. Therefore, j + an irrational number must be irrational also.