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  4. The number of non-negative integral values of k for which the equation 5 x2+(13-k) x-3 k-6=0 has atleast one real solution in (-2,2) are

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

Complex Numbers and Quadratic Equations

Solution:

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