Question

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

• A. 1
• B. 2
• C. 5
• D. 8

Answer 1

Answer: (B)

Given, $-3x^{4}$+det $\left[\begin{matrix}1&x&x^{2}\\ 1&x^{2}&x^{4}\\ 1&x^{3}&x^{6}\end{matrix}\right]=0$
$\Rightarrow x^{8}+x^{5}+x^{5}-x^{4}-x^{7}-x^{7}=3x^{4}$
$\Rightarrow x^{8}-2x^{7}+2x^{5}-4x^{4}=0$
$\Rightarrow x^{4}\left[x^{4}-2x^{3}+2x-4\right]=0$
$\Rightarrow x^{4}\left[x^{3}\left(x-2\right)+2\left(x-2\right)\right]=0$
$\Rightarrow x^{4}\left(x^{3}+2\right)\left(x-2\right)=0$
$\therefore $ $x$ is an integer, so $x=0,2$