Question

The set of all values of m for which both the roots of the equation $x^2(m+1)x+m+4=0$are real and negative, is

• A. $(-\infty ,-3]\cup [5,\infty )$
• B. $[-3,5]$
• C. $(-4,-3]$
• D. $(-3,-1]$

Answer 1

If the equation has real roots, then the discriminant must be positive. Let us verify : $d=b^2-4ac>0$ $\Rightarrow (m+1{)}^2-4\times 1\times (m+4)>0\Rightarrow m^2+2m+1-4m-16>0\Rightarrow m^2-2m-15>0\Rightarrow (m-5\left)\right(m+3)>0\Rightarrow m-5>0andm+3>0orm-5<0andm+3<0\Rightarrow m>5andm>-3orm<5andm<-3\Rightarrow mbelongsto(5,\infty )\cup (-\infty ,-3)...(i)\text{Now in order for the roots to be negative then}{x}_1+x_2<0=\frac{-b}{a}=(m+1)<0\Rightarrow m<-1Also{x}_1\times x_2=\frac{c}{a}=m+4>0\Rightarrow m>-4\Rightarrow mbelongsto(-\infty ,-1)\cup (-4,\infty )...(ii\left)From\right(i\left)and\right(ii\left)weconcludembelongsto\right(-\infty ,-3)\cup (5,\infty )$