Question

please solve question number 12 .​

Answer 1

Given,

$\longrightarrow\left|\begin{array}{ccc}7&6&x\\2&x&2\\x&3&7\end{array}\right|=0$

Perform the operation $C_1\to C_1-C_3.$

$\longrightarrow\left|\begin{array}{ccc}7-x&6&x\\2-2&x&2\\x-7&3&7\end{array}\right|=0$

$\longrightarrow\left|\begin{array}{ccc}-(x-7)&6&x\\0&x&2\\x-7&3&7\end{array}\right|=0$

We can take $x-7$ outside from $C_1.$

$\longrightarrow(x-7)\left|\begin{array}{ccc}-1&6&x\\0&x&2\\1&3&7\end{array}\right|=0$

This implies,

$\longrightarrow\underline{\underline{x=7}}$

Or,

$\longrightarrow\left|\begin{array}{ccc}-1&6&x\\0&x&2\\1&3&7\end{array}\right|=0$

Expanding this discriminant along $C_1,$

$\longrightarrow-1(7x-6)+1(12-x^2)=0$

$\longrightarrow x^2+7x-18=0$

$\longrightarrow x^2+9x-2x-18=0$

$\longrightarrow x(x+9)-2(x+9)=0$

$\longrightarrow (x+9)(x-2)=0$

This implies,

$\longrightarrow\underline{\underline{x=-9}}$

$\longrightarrow\underline{\underline{x=2}}$

So the roots of the equation are,

$\longrightarrow \underline{\underline{x\in\{-9,\ 2,\ 7\}}}$

And the other roots are 2 and 7.