Question

If both roots of the equation $x^2-(m+1)x+(m+4)=0$ are negative, then $m$ equals

• A. $-7,m<-5$
• B. $-4<m<-1$
• C. $2,m<5$
• D. $Noneofthese$

Answer 1

Answer: (B)

$-4<m<-1$

Given $x^2-(m+1)x+m+4=0$,

If the equation has real roots, then

$b^2-4ac>0$

$\Rightarrow (m+1{)}^2-4\times 1\times (m+4)>0$

$\Rightarrow m^2+1+2m-4m-16>0$

$\Rightarrow m^2-2m-15>0$

$\Rightarrow (m-5)(m+3)>0$

$\Rightarrow m-5>0andm+3>0$

$\Rightarrow m>5andm>-3$

Now for roots to be negative then,

$x_1+x_2<0and{x}_1{x}_2>0$

$x_1+x_2=\frac{-b}{a}=(m+1)$

$m+1<0$

$m<-1$

$x_1{x}_2=\frac{c}{a}=(m+4)$

$m+4>0$

$m>-4$

According to the question both the roots are negative so $m$ belongs to

$m<-1$ and $m>-4$