Question

The values of $m$ for which roots of the quadratic equation $-x^2+(2m+3)x-m^2=0$ lies on either side of $3.$

• A. $(0,6)$
• B. $[\frac{-3}{4},0]$
• C. $(-6,0)$
• D. $[\frac{-3}{4},\infty ]$

Answer 1

The correct option is A $(0,6)$

Let $f\left(x\right)=-x^2+(2m+3)x-m^2$

On comparing with standard quadratic equation $y=a{x}^2+bx+c$ we get $a=-1,b=2m+3,c=-m^2.$

Since $a<0$
$\Rightarrow \text{Downward opening parabola}$

$\therefore$ Required conditions are
$\left(i\right)D>0$ (Since there are 2 unequal roots)
$D=b^2-4ac>0$
$D=(2m+3{)}^2-4.(-1).(-m^2)>0$
$\Rightarrow 12m+9>0$
$\Rightarrow m>\frac{-3}{4}$
​​​​$\Rightarrow m\in (\frac{-3}{4},\infty )\cdots \left(1\right)$

$\left(ii\right)f\left(3\right)>0$
$-(3{)}^2+(2m+3\left)\right(3)-m^2>0$
$\Rightarrow m(m-6)<0$
$\Rightarrow m\in (0,6)\cdots \left(2\right)$

Final set of solution will be the intersection of $1,2.$
$m\in (\frac{-3}{4},\infty )\cap (0,6)$

$m\in (0,6)$