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(B) Given the determinant equation: $\left| \begin{array}{ccc} 1 & 4 & 20 \ 1 & -2 & 5 \ 1 & 2x & 5x^2 \end{array} ight| = 0$ Applying row operati

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$ -1, 2 $

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(B) Given the determinant equation: $\left| \begin{array}{ccc} 1 & 4 & 20 \ 1 & -2 & 5 \ 1 & 2x & 5x^2 \end{array} ight| = 0$ Applying row operations $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$: $\left| \begin{array}{ccc} 0 & 6 & 15 \ 0 & -2-2x & 5-5x^2 \ 1 & 2x & 5x^2 \end{array} ight| = 0$ Taking common factors $3$ from $R_1$ and $5$ from $R_2$: $15 \left| \begin{array}{ccc} 0 & 2 & 5 \ 0 & -2(1+x) & 5(1-x)(1+x) \ 1 & 2x & 5x^2 \end{array} ight| = 0$ Taking $(1+x)$ common from $R_2$: $15(1+x) \left| \begin{array}{ccc} 0 & 2 & 5 \ 0 & -2 & 5(1-x) \ 1 & 2x & 5x^2 \end{array} ight| = 0$ Expanding along the first column: $15(1+x) [1(2 	imes 5(1-x) - 5 	imes (-2))] = 0$ $15(1+x) [10-10x + 10] = 0$ $15(1+x) [20-10x] = 0$ $150(1+x)(2-x) = 0$ Thus,$x = -1$ or $x = 2$.

The roots of the equation $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$ are

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