Q45 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

Sum of the positive roots of the equation $\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 2 x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$

  • Correct answer
    A
    $\frac{1+\sqrt{13}}{2}$
  • B
    $1$
  • C
    $\frac{\sqrt{13}-1}{2}$
  • D
    $3$

Solution

$\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 2 x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$ $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$ $\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 1-x^2 & -3 & 0 \\ -x^2-x+2 & -x-3 & 0\end{array}\right|=0$ $\Rightarrow\left(1-x^2\right)(-x-3)+3\left(-x^2-x+2\right)=0$ $\Rightarrow x^3-4 x+3=0 \Rightarrow(x-1)\left(x^2+x-3\right)=0$ $\Rightarrow x=1, \frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}$ Sum of positive root $=1+\frac{\sqrt{13}-1}{2}=\frac{1+\sqrt{13}}{2}$

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

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