Question

The solution set of $x^2-4x+1<0$ is

• A. $\varphi \left(\text{empty set}\right)$
• B. $R$
• C. $[0,\infty )$
• D. $(2-\sqrt{3},2+\sqrt{3})$

Answer 1

The correct option is D $(2-\sqrt{3},2+\sqrt{3})$

$x^2-4x+1<0$
Roots are
$\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\therefore x\in (2-\sqrt{3},2+\sqrt{3})$