Multiple choice

If one root of the quadratic equation $x^2+px+q=0$ is $2-\sqrt{3}$; where p, q $\in$ Q. Then which of the following is true?

  1. $p^2-4q+12=0$
  2. $p^2-4q-12=0$
  3. $q^2-4p+12=0$
  4. $q^2-4p-12=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the coefficients p and q are rational, if 2-sqrt(3) is a root, then its conjugate 2+sqrt(3) must also be a root. The sum of roots is -p = (2-sqrt(3)) + (2+sqrt(3)) = 4, so p = -4. The product of roots is q = (2-sqrt(3))(2+sqrt(3)) = 4-3 = 1. Substituting into p^2-4q-12 gives (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0.

AI explanation

Since p and q are rational numbers and one root is 2 - sqrt(3), the other root must be its conjugate, 2 + sqrt(3). The sum of the roots is (2 - sqrt(3)) + (2 + sqrt(3)) = 4, so p = -4. The product of the roots is (2 - sqrt(3))(2 + sqrt(3)) = 4 - 3 = 1, so q = 1. Substituting these values into p^2 - 4q gives 16 - 4(1) = 12, therefore p^2 - 4q - 12 = 0.

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