Question

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

• A. $x^2-8x+13=0$
• B. $x^2-8x+19=0$
• C. $x^2-8x-13=0$
• D. $x^2-8x-19=0$

Answer 1

The correct option is B $x^2-8x+19=0$

The roots must exist in conjugate pairs since the coefficients are real.

Hence, if one root of the quadratic equation is $4-i\sqrt{3}$ then the other root will be $4+i\sqrt{3}$.

Hence the quadratic equation will be

$(x-(4+i\sqrt{3}\left)\right)(x-(4-i\sqrt{3}\left)\right)=0$

$x^2-x(4+i\sqrt{3}+4-i\sqrt{3})+\left[\right(4-i\sqrt{3})\times (4+i\sqrt{3}\left)\right]=0$

$x^2-8x+(16+3)=0$

$x^2-8x+19=0$