Question

The value of $m$ for which atleast one root of the quadratic equation $(m^2+m+2)x^2-(m+5)x+2=0$ is greater than $-1$ is

• A. $[-1,-\frac{7}{9}]$
• B. $[-\frac{7}{9},1]$
• C. $[-1,\frac{7}{9}]$
• D. $[\frac{7}{9},1]$

Answer 1

The correct option is C $[-1,\frac{7}{9}]$

Given: $(m^2+m+2)x^2-(m+5)x+2=0;$
have atleast one root is greater than $-1.$

$\text{let}y=(m^2+m+2)x^2-(m+5)x+2$
On comparing with standard form of quadratic equation $y=a{x}^2+bx+c$ we get $a=m^2+m+2,b=-(m+5),c=2.$
$\text{Taking}a=m^2+m+2=(m+\frac{1}{2}{)}^2+\frac{3}{2}$
$\Rightarrow a>0$, hence we get an upward facing parabola

Now we will have $3$ cases.
Case $1:$ Both the roots are greater than $-1.$

Required conditions are
$\left(i\right)D\ge 0$
$D=b^2-4ac\ge 0$
$D=(-(m+5){)}^2-4.(m^2+m+2).(2)\ge 0$
$\Rightarrow 7m^2-2m-9\le 0$
$\Rightarrow 7m^2+7m-9m-9\le 0$
$\Rightarrow 7m(m+1)-9(m+1)\le 0$
$\Rightarrow (7m-9)(m+1)\le 0$
$\Rightarrow m\in [-1,\frac{7}{9}]\cdots \left(1\right)$

$\left(ii\right)f(-1)>0$
$(m^2+m+2)(-1{)}^2-(m+5\left)\right(-1)+2>0$
$\Rightarrow m^2+2m+9>0$
$\Rightarrow m\in R\cdots \left(2\right)$

$\left(iii\right)x$ coordinate of the vertex will be greater than $-1$
$\therefore \frac{-b}{2a}>-1$
$\frac{-(-(m+5\left)\right)}{2(m^2+m+2)}>-1$
$\Rightarrow 2m^2+3m+9>0)$
$\Rightarrow m\in R\cdots \left(3\right)$

From equation $\left(1\right),\left(2\right),\left(3\right)$
$m\in [-1,\frac{7}{9}]\cdots \left(4\right)$

Case $2:$ One root is greater than $-1$ and other is less than $-1.$

Required condition is
$\left(i\right)f(-1)<0$
$(m^2+m+2)(-1{)}^2-(m+5\left)\right(-1)+2<0$
$\Rightarrow m^2+2m+9<0$
$\Rightarrow m\in \varphi \cdots \left(5\right)$

Case $3:$ One root is greater than $-1$ and other root (smaller) is equal to $-1.$

Required conditions are
$\left(i\right)D\ge 0$
$D=b^2-4ac\ge 0$
$D=(-(m+5){)}^2-4.(m^2+m+2).(2)\ge 0$
$\Rightarrow 7m^2-2m-9\le 0$
$\Rightarrow 7m^2+7m-9m-9\le 0$
$\Rightarrow 7m(m+1)-9(m+1)\le 0$
$\Rightarrow (7m-9)(m+1)\le 0$
$\Rightarrow m\in [-1,\frac{7}{9}]\cdots \left(6\right)$

$\left(ii\right)f(-1)=0$
$(m^2+m+2)(-1{)}^2-(m+5\left)\right(-1)+2=0$
$\Rightarrow m^2+2m+9=0$
$\Rightarrow m\in \varphi \cdots \left(7\right)$

$\left(iii\right)x$ coordinate of the vertex will be greater than $-1$
$\therefore \frac{-b}{2a}>-1$
$\frac{-(-(m+5\left)\right)}{2(m^2+m+2)}>-1$
$\Rightarrow 2m^2+3m+9>0)$
$\Rightarrow m\in R\cdots \left(8\right)$


From equation $\left(6\right),\left(7\right),\left(8\right)$
$\Rightarrow m\in \varphi \cdots \left(9\right)$

Final solution set will be the union of the final results of cases $1,2,3.$
$m\in [-1,\frac{7}{9}]\cup \varphi \cup \varphi$

$\Rightarrow m\in [-1,\frac{7}{9}]$