Sum of the positive roots of the equation lef | vedclass

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(A) Given the determinant equation: $\left|\begin{array}{ccc}x^2+2x & x+2 & 1 \ 2x+1 & x-1 & 1 \ x+2 & -1 & 1\end{array}ight|=0$ Applying row oper

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$\frac{1+\sqrt{13}}{2}$

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(A) Given the determinant equation: $\left|\begin{array}{ccc}x^2+2x & x+2 & 1 \ 2x+1 & x-1 & 1 \ x+2 & -1 & 1\end{array}ight|=0$ Applying row operations $R_2 ightarrow R_2-R_1$ and $R_3 ightarrow R_3-R_1$: $\left|\begin{array}{ccc}x^2+2x & x+2 & 1 \ 1-x^2 & -3 & 0 \ -x^2-x+2 & -x-3 & 0\end{array}ight|=0$ Expanding along the third column: $1 \cdot [(1-x^2)(-x-3) - (-3)(-x^2-x+2)] = 0$ $(1-x^2)(-x-3) + 3(-x^2-x+2) = 0$ $-x-3+x^3+3x^2-3x^2-3x+6 = 0$ $x^3-4x+3 = 0$ Factoring the cubic equation: $(x-1)(x^2+x-3) = 0$ The roots are $x=1$ and $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-3)}}{2} = \frac{-1 \pm \sqrt{13}}{2}$. The positive roots are $x=1$ and $x = \frac{\sqrt{13}-1}{2}$. Sum of positive roots $= 1 + \frac{\sqrt{13}-1}{2} = \frac{2+\sqrt{13}-1}{2} = \frac{1+\sqrt{13}}{2}$.

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

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