Let ∆ = x-6-12-3xx-3-32xx+2Applying R1 = R1 - R2, we get∆ = x-23x-2-x-22-3xx-3-32xx+2⇒∆=x-213-12-3xx-3-32xx+2Applying C1 = C1 + C3 and C2 = C2 + 3C3∆ = x-200-1x-1-9x-3x-15x+6x+2⇒∆ = x-2x-100-11-9x-315x+6x+2⇒∆ = x-2x-1 × 15x+6+9⇒∆ = x-2x-15x+15Now, ∆ = 0⇒ x-2x-15x+15 = 0⇒x = 2; x = 1; x = -3So, x = 2 is a root of the given equation

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Let ∆ = |\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| Applying R_{1} = R_{1} - R_{2}, we get ∆ = |\begin{matrix} x - 2 & 3 (x - 2) & - (x - 2) \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| \Rightarrow ∆ = (x - 2) |\begin{matrix} 1 & 3 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| Applying C_{1} = C_{1} + C_{3} and C_{2} = C_{2} + 3 C_{3} ∆ = (x - 2) |\begin{matrix} 0 & 0 & - 1 \\ x - 1 & - 9 & x - 3 \\ x - 1 & 5 x + 6 & x + 2 \end{matrix}| \Rightarrow ∆ = (x - 2) (x - 1) |\begin{matrix} 0 & 0 & - 1 \\ 1 & - 9 & x - 3 \\ 1 & 5 x + 6 & x + 2 \end{matrix}| \Rightarrow ∆ = (x - 2) (x - 1) \times 1 [5 x + 6 + 9] \Rightarrow ∆ = (x - 2) (x - 1) (5 x + 15) Now, ∆ = 0 \Rightarrow (x - 2) (x - 1) (5 x + 15) = 0 \Rightarrow x = 2; x = 1; x = - 3 So, x = 2 is a root of the given equation .

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