Question

If both the roots of the quadratic equation $x^2-2kx+k^2+k-5=0$ are less than $5$, then $k$ lies in the interval

• A. $(-\infty ,4)$
• B. $(6,\infty )$
• C. $[4,5]$
• D. $(5,6]$

Answer 1

The correct option is A $(-\infty ,4)$

$f\left(x\right)=x^2-2kx+k^2+k-5$
both roots are less than $5$, then
$\left(i\right)$ Discriminant $\ge 0$
$\left(ii\right)f\left(5\right)>0\left(iii\right)\frac{Sumofroots}{2}<5$

Hence $\left(i\right)4k^2-4\left(1\right)(k^2+k-5)\ge 04k^2-4k^2-4k+20\ge 04k\le 20\Rightarrow k\le 5$

$\left(ii\right)f\left(5\right)>0;25-10k+k^2+k-5>0or{k}^2-9k+20>0ork(k-4)-5(k-4)>0or(k-5)(k-4)>0\Rightarrow k\in (-\infty ,4)\cup (5,\infty )$

$\left(iii\right)\frac{Sumofroots}{2}=-\frac{b}{2a}=\frac{2k}{2}<5$

The intersection of $\left(i\right),\left(ii\right)\&\left(iii\right)$ gives $k\in (-\infty ,4).$