Question

The equation $(x^2-5x+1)(x^2+x+1)+8x^2=0$ has

• A. four real and distinct roots
• B. three real and distinct roots
• C. two real and distinct roots
• D. only one real root

Answer 1

The correct option is C two real and distinct roots

$(x^2-5x+1)(x^2+x+1)+8x^2=0$
$\Rightarrow (x+\frac{1}{x}-5)(x+\frac{1}{x}+1)+8=0$
Let $x+\frac{1}{x}=t$
Then, $(t-5)(t+1)+8=0$
$\Rightarrow t^2-4t-5+8=0$
$\Rightarrow t^2-4t+3=0$
$\Rightarrow t=1$ or $t=3$
But $x+\frac{1}{x}=1$ rejected because $x+\frac{1}{x}\in (-\infty ,-2]\cup [2,\infty )$
So, $x+\frac{1}{x}=3$
$\Rightarrow x^2-3x+1=0$
$D=9-4=5>0$
$\Rightarrow$ Two real and distinct roots.