Multiple choice

If $4-i\sqrt{3}$ is a root of quadratic equation with real coefficients then the equation is?

  1. $x^2-8x+13=0$
  2. $x^2-8x+19=0$
  3. $x^2-8x-13=0$
  4. $x^2-8x-19=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If a quadratic equation has real coefficients and a complex root 4-i*sqrt(3), the other root must be the conjugate 4+i*sqrt(3). The equation is (x - (4-i*sqrt(3))) * (x - (4+i*sqrt(3))) = 0. This is (x-4)^2 - (i*sqrt(3))^2 = 0 -> x^2 - 8x + 16 + 3 = 0 -> x^2 - 8x + 19 = 0.

AI explanation

For a quadratic equation with real coefficients, any complex root must occur in conjugate pairs. Therefore, the other root of the equation is 4 + i*sqrt(3). The sum of the roots is (4 - i*sqrt(3)) + (4 + i*sqrt(3)) = 8, making the coefficient of x equal to -8. The product of the roots is (4 - i*sqrt(3))*(4 + i*sqrt(3)) = 16 - 3i^2 = 19, which equals the constant term, yielding the equation x^2 - 8x + 19 = 0.

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