Q53 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

If $\mathrm{P}(\mathrm{x})=0$ is a polynomial equation of least degree with integer coefficients and $\sqrt{2}+\sqrt{3} i$ is one of its roots, then that equation is

  • A
    $\mathrm{x}^6-2 \mathrm{x}^4+2 \mathrm{x}^2-25=0$
  • B
    $x^5+3 x^4+2 x^2+24=0$
  • Correct answer
    C
    $x^4+2 x^2+25=0$
  • D
    $x^4-2 x^2+25=0$

Solution

$(\sqrt{2}+\sqrt{3} i)$ is a root of $\mathrm{P}(\mathrm{x})$ then $(\sqrt{2}-\sqrt{3} \mathrm{i})$, $(-\sqrt{2}+\sqrt{3} i)$ and $(-\sqrt{2}-\sqrt{3} i)$ will also be roots of $p(x)$ $$ \begin{aligned} & \therefore(x-(\sqrt{2}+\sqrt{3} i))(x-(\sqrt{2}-\sqrt{3} i))(x-(-\sqrt{2}+\sqrt{3} i)) \\ & (x-(-\sqrt{2}-\sqrt{3} i))=0 \\ & \Rightarrow\left(x^2-2 \sqrt{2} x+5\right)\left(x^2+2 \sqrt{2} x+5\right)=0 \\ & \Rightarrow\left(x^2+5\right)^2-(2 \sqrt{2} x)^2=0 \\ & \Rightarrow x^4+25+10 x^2-8 x^2=0 \\ & \Rightarrow x^4+2 x^2+25=0 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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