Question

For what value of $k$, are the roots of the quadratic equation $k{x}^2+(k-2)x+6=0$ equal?

Answer 1

$k{x}^2+(k-2)x+6=0$

Roots are equal$\Rightarrow$ Discriminant$=0$

$\Delta ={(k-2)}^2-4\times k\times 6=0$

$\Rightarrow k^2-4k+4-24k=0$

$\Rightarrow k^2-28k+4=0$

$\Rightarrow k^2-2\times 14k+{14}^2-{14}^2+4=0$

$\Rightarrow k^2-2\times 14k+196-196+4=0$

$\Rightarrow {(k-14)}^2=192$

$\Rightarrow k-14=\pm \sqrt{192}=\pm \sqrt{4\times 4\times 4\times 3}=\pm 8\sqrt{3}$

$\therefore k=14\pm 8\sqrt{3}$

Hence $k=14+8\sqrt{3},14-8\sqrt{3}$