Question

The number of integers $x$ satisfying
$-3x^4+\text{det}\left[\begin{array}{ccc}1& x& x^2\\ 1& x^2& x^4\\ 1& x^3& x^6\end{array}\right]=0$
is equal to

• A. $1$
• B. $2$
• C. $5$
• D. $8$

Answer 1

The correct option is B $2$

$-3x^4+\text{det}\left[\begin{array}{ccc}1& x& x^2\\ 1& x^2& x^4\\ 1& x^3& x^6\end{array}\right]=0$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 1& x^2& x^4\\ 1& x^3& x^6\end{array}\right|=3x^4$
Using row opperations,
$R_2\to R_2-R_1$ and
$R_3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 0& x^2-x& x^4-x^2\\ 0& x^3-x& x^6-x^2\end{array}\right|=3x^4$
$\Rightarrow (x^2-x)(x^6-x^2)-(x^4-x^2)(x^3-x)=3x^4\Rightarrow (x^8-x^4-x^7+x^3)-(x^7-x^5-x^5+x^3)=3x^4\Rightarrow x^8-2x^7+2x^5-4x^4=0\Rightarrow x^7(x-2)+2x^4(x-2)=0\Rightarrow (x-2)(x^7+2x^4)=0\Rightarrow x^4(x-2)(x^3+2)=0$
Integer solutions,
$x=0,2$
NUmber of integer solutions $=2$