Question

if 4-i√3 root of quadratic equation with real coefficients then the equation is

Answer 1

Answer:

$\sf{The \ required \ quadratic \ equation \ is}$

$\sf{x^{2}-8x+19=0.}$

Given:

$\sf{If \ 4-i\sqrt3 \ is \ root \ of \ quadratic \ equation}$

$\sf{with \ real \ coefficients.}$

To find:

$\sf{The \ equation.}$

Solution:

$\sf{If \ roots \ of \ quadratic \ equation \ are}$

$\sf{in \ complex \ number \ they \ are \ in}$

$\sf{conjugate \ pair.}$

$\sf{Conjugate \ of \ 4-i\sqrt3 \ is \ 4+i\sqrt3}$

$\sf{\therefore{Roots \ are \ 4-i\sqrt3 \ and \ 4+i\sqrt3}}$

$\sf{Sum \ of \ roots=(4-i\sqrt3)+(4+i\sqrt3)}$

$\sf{\therefore{Sum \ of \ roots=8...(1)}}$

$\sf{Product \ of \ roots=(4-i\sqrt3)(4+i\sqrt3)}$

$\sf{\therefore{Product \ of \ roots=4^{2}-(i\sqrt3)^{2}}}$

$\sf{\therefore{Product \ of \ roots=16-3i^{2}}}$

$\sf{...[But \ i^{2}=-1]}$

$\sf{\therefore{Product \ of \ roots=16+3}}$

$\sf{\therefore{Product \ of \ roots=19...(2)}}$

$\sf{Quadratic \ equation \ can \ be \ written \ as}$

$\sf{x^{2}-(Sum \ of \ roots)x+(Product \ of \ roots)=0}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\leadsto{x^{2}-8x+19=0}}$

$\sf\purple{\tt{\therefore{The \ required \ quadratic \ equation \ is}}}$

$\sf\purple{\tt{x^{2}-8x+19=0.}}$