Question

$\mathrm{If}\left(\begin{array}{ccc}{x}^2& x& 1\\ 0& 2& 1\\ 3& 1& 4\end{array}\right|=28,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{is}$

• A. $x=2$
• B. $x=3$
• C. $x=-\frac{17}{7}$
• D. $x=-\frac{17}{2}$

Answer 1

Answer: (A), (C)

The correct options are
A $x=2$
C $x=-\frac{17}{7}$

On expanding along second row, we get:
$0\left(4x-1\right)+2\left(4x^2-3\right)-1\left(x^2-3x\right)=28\Rightarrow 8x^2-6-x^2+3x-28=0\Rightarrow 7x^2+3x-34=0\Rightarrow 7x^2-14x+17x-34=0\Rightarrow \left(x-2\right)\left(7x+17\right)=0\Rightarrow x=2\mathrm{and}x=-\frac{17}{7}$