Question

For what values of $k$ will the quadratic equation : ${2x}^2-kx+1=0$ have real and equal roots?

• A. $\pm 2\sqrt{2}$
• B. $\pm \sqrt{2}$
• C. $\pm 3\sqrt{2}$
• D. $\pm \sqrt{3}$

Answer 1

The correct option is A $\pm 2\sqrt{2}$

${2x}^2-kx+1=0$
Hence, $a=2,b=-k.c=1$
$D=b^2-4ac={(-k)}^2-4\times 2\times 1-k^2-8$
We know that a quadratic equation has real and equal roots, if
$D=0\Leftarrow k^2-8=0\Rightarrow k^2=8$
$k=\pm \sqrt{8}=\pm 2\sqrt{2}$