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### Answers (1)

Answer: $x=1,1,-9$

Hint: Use determinant formula

Given: $\left|\begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array}\right|=0$

Solution:

\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|=0 \end{aligned}

\begin{aligned} &(x+9) \text { common from } \mathrm{C}_{1}\\ &=(x+9)\left|\begin{array}{ccc} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{array}\right|\\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(x+9)\left|\begin{array}{ccc} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &(x+9)(x-1)^{2}=0\\ &x+9=0,(x-1)^{2}=0\\ &x=-9, x=1,1\\ &x=1,1,-9 \end{aligned}

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