Question

The number of non-negative integral values of $k$ for which the equation $5x^2+(13-k)x-3k-6=0$ has atleast one real solution in $(-2,2)$ are

• A. 21
• B. 20
• C. 7
• D. 8

Answer 1

Answer: (D)

$5x^2+(13-k)x-(3k+6)=0$
$x=\frac{(k-13)\pm \sqrt{(13-k{)}^2+20(3k+6)}}{10}$
$x=\frac{(k-13)\pm \sqrt{k^2-26k+169+60k+120}}{10}$
$x=\frac{(k-13)\pm (k+17)}{10}\Rightarrow x=-3,\frac{2k+4}{10}$
$\text{ Since, }x\in (-2,2)$
$\therefore -2<\frac{2k+4}{10}<2$
$-20<2k+4<20$
$-24<2k<16$
$-12<k<8$