• question_answer Number of possible integral values of m in $(-10,10)$ for which the quadratic equation ${{x}^{2}}+(m+6)\left| x \right|+2m+8=0$ has two distinct real solutions are A)  1                     B)  3 C)  5                                 D)  8

$|x|\,=\frac{-(m+6)\,\pm \,(m+2)}{2}\,(reject+sign)$ $|x|=-(m+4)$ For two distinct solution, m + 4 < 0 $\Rightarrow \,m<-4$ $\Rightarrow$ Number of integral values of m in (-10, 10] are {-9, -8, -7, -6, -5} i.e., 5 values Alternating: Since equation has two distinct solution and therefore product of the roots must be less than zero. $\Rightarrow \,2(m+4)<0\Rightarrow \,m<-4$. $f(\lambda )\,=\det \,(A-\lambda I)\,=\left| \begin{matrix} 1-\lambda & 2 \\ -1 & 3-\lambda \\ \end{matrix} \right|\,=(1-\lambda )\,(3-\lambda )\,+2$$={{\lambda }^{2}}-4\,\lambda +5={{(\lambda -2)}^{2}}+1$ Clearly, ${{f}_{\min }}\,(\lambda =2)=1$