If x=k satisfies the equation left|begin{arra | vedclass

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(D) Given the determinant equation: $\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \ x-4 & 3(x-3) & 5(x-5) \ x-8 & 3(x-9) & 5(x-25)\end{array}ight

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$x^2+2x-3=0$

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(D) Given the determinant equation: $\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \ x-4 & 3(x-3) & 5(x-5) \ x-8 & 3(x-9) & 5(x-25)\end{array}ight|=0$. Taking common factors $3$ and $5$ from $C_2$ and $C_3$ respectively: $15 \left|\begin{array}{ccc}x-2 & x-1 & x-1 \ x-4 & x-3 & x-5 \ x-8 & x-9 & x-25\end{array}ight|=0$. Applying $C_2 ightarrow C_2 - C_3$: $\left|\begin{array}{ccc}x-2 & 0 & x-1 \ x-4 & 2 & x-5 \ x-8 & 16 & x-25\end{array}ight|=0$. Expanding along $C_2$: $-2((x-2)(x-25) - (x-1)(x-8)) + 16((x-2)(x-5) - (x-1)(x-4)) = 0$. $-2(x^2-27x+50 - (x^2-9x+8)) + 16(x^2-7x+10 - (x^2-5x+4)) = 0$. $-2(-18x+42) + 16(-2x+6) = 0$. $36x - 84 - 32x + 96 = 0 \Rightarrow 4x + 12 = 0 \Rightarrow x = -3$. Checking the options for $x = -3$: For $x^2+2x-3=0$, $(-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0$. Thus, $x = -3$ satisfies $x^2+2x-3=0$.

If $x=k$ satisfies the equation $\left|\begin{array}{ccc}x-2 & 3x-3 & 5x-5 \\ x-4 & 3x-9 & 5x-25 \\ x-8 & 3x-27 & 5x-125\end{array}\right|=0$, then $x=k$ also satisfies the equation

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