Question

If $ 4-i\sqrt{3} $ is a root of the quadratic equation having all 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

Hint:Use the fact that if a polynomial equation(with real coefficients) has a complex root, then the complex conjugate of the root is also the solution of the equation. Hence if $ a+ib $ is a root of a polynomial equation, then $ a-ib $ is also the root of the equation. Use the fact that a quadratic equation whose sum of roots is “a” and product of roots is “b” is $ {{x}^{2}}-ax+b $ . Hence find the required equation.

Complete step-by-step answer:
We know that if a polynomial equation(with real coefficients) has a complex root, then the complex conjugate of the root is also the solution of the equation. Since the quadratic equation has $ 4-i\sqrt{3} $ as one of the roots, the other root of the equation is $ \overline{4-i\sqrt{3}}=4+i\sqrt{3} $
Hence, we have
Sum of roots $ =4+i\sqrt{3}+4-i\sqrt{3}=8 $
Product of roots $ =\left( 4+i\sqrt{3} \right)\left( 4-i\sqrt{3} \right)={{4}^{2}}-{{\left( i\sqrt{3} \right)}^{2}}={{4}^{2}}+3=19 $
We know that a quadratic equation whose sum of roots is “a” and product of roots is “b” is $ {{x}^{2}}-ax+0 $ .
Hence the required quadratic equation is $ {{x}^{2}}-8x+19=0 $
Hence option [b] is correct.

Note: Verification:
We can verify the correctness of our solution by checking that $ 4-i\sqrt{3} $ is a root of the equation.
We have
 $ {{\left( 4-i\sqrt{3} \right)}^{2}}=16-3-8\sqrt{3}i=13-8\sqrt{3} $
Hence, we have
 $ \begin{align}
  & {{x}^{2}}-8x+9={{\left( 4-i\sqrt{3} \right)}^{2}}-8\left( 4-i\sqrt{3} \right)+19 \\
 & =13-8\sqrt{3}i-32+8i\sqrt{3}+19 \\
 & =0 \\
\end{align} $
Hence our solution is verified to be correct.