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Please Solve R. D. Sharma class 12 Chapter determinants Exercise  5.2 Question 51 Maths textbook Solution.

Answers (1)

Answer: $$ x=1,2,-3

Hint Use determinant formula

Given: $$ \text { S.T } x=2 \text { is root of equation }\left|\begin{array}{ccc} x & -6 & -1 \\ 2 & 3 x & x-3 \\ -3 & 2 x & x+3 \end{array}\right|=0

Solution:

$$ \left|\begin{array}{ccc} x & -6 & -1 \\ 2 & 3 x & x-3 \\ -3 & 2 x & x+3 \end{array}\right|=0

$$ \begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &=x\left(-3 x^{2}-6 x-2 x^{2}+6 x\right)+(2 x+4+3 x-9)+30 x-30+5 x=0\\ &=-5 x^{2}+30 x-30+5 x=0\\ &=-5\left(x^{3}+7 x+6\right)=0\\ &=x^{3}+7 x+6=0\\ &=(x-1)\left(x^{2}+x-6\right)=0\\ &=(x-1)\left(x^{2}+3 x-2 x-6\right)=0\\ &=(x-1)(x+3)(x-2)=0\\ &x=1,2,-3 \end{aligned}

 

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