Question

If $\left[\begin{array}{ccc}x& 4& -1\end{array}\right]\left[\begin{array}{ccc}2& 1& 0\\ 1& 0& 2\\ 0& 2& 4\end{array}\right]\left[\begin{array}{c}x\\ 4\\ -1\end{array}\right]=0$ then $x$ is equal to

• A. $-1+\sqrt{6}$
• B. $8\pm \sqrt{5}$
• C. $-2\pm \sqrt{10}$
• D. $3\pm \sqrt{6}$

Answer 1

Answer: (C)

$-2\pm \sqrt{10}$

${\left[\begin{array}{ccc}x& 4& 1\end{array}\right]}_{1\times 3}{\left[\begin{array}{ccc}2& 1& 0\\ 1& 0& 2\\ 0& 2& 4\end{array}\right]}_{3\times 3}{\left[\begin{array}{c}x\\ 4\\ -1\end{array}\right]}_{3\times 1}=0$

${\left[\begin{array}{ccc}x& 4& 1\end{array}\right]}_{1\times 3}{\left[\begin{array}{c}2x+4+0\\ x+0-2\\ 0+8-4\end{array}\right]}_{3\times 1}=0$

$\Rightarrow 2x^2+4x+4x-8-4=0$

$\Rightarrow 2x^2+8x-12=0$

$\Rightarrow x^2+4x-6=0$

$\Rightarrow x^2+2\times 2x+4-4-6=0$

$\Rightarrow {(x+2)}^2=10$

$\Rightarrow x+2=\pm \sqrt{10}$

$\therefore x=-2\pm \sqrt{10}$