Question

The values of $m$ for which the quadratic expression $y=(m^2-9)x^2+(m-3)x-7$ is always greater than or equal to $-1$ is

• A. $[-\frac{25}{9},3]$
• B. $\varphi$
• C. $R$
• D. $(-\infty ,-3)\cup (3,\infty )$

Answer 1

The correct option is B $\varphi$

Given: $y=(m^2-9)x^2+(m-3)x-7,y\ge -1$

On comparing with standard quadratic expression $y=a{x}^2+bx+c$,we get $a=m^2-9,b=m-3,c=-7.$

Here, we have given that the given equation is always greater than or equal to $-1,$ which means the minimum value of the expression is $-1,$ and the parabola is opening upward.

$\therefore a>0\Rightarrow m^2-9>0$
$\Rightarrow (m+3)(m-3)>0$
$\Rightarrow m\in (-\infty ,-3)\cup (3,\infty )$

And, $\frac{-D}{4a}\ge -1$
$\frac{-(b^2-4ac)}{4a}\ge -1$

$\frac{-\left(\right(m-3{)}^2-4(m^2-9)(-7))}{2(m^2-9)}\ge -1$
$\Rightarrow \frac{(m^2-6m+9+28m^2-252)}{2(m^2-9)}\le 1$
$\Rightarrow 29m^2-6m-243\le 2(m^2-9)$
$\Rightarrow 27m^2-6m-225\le 0$
$\Rightarrow 9m^2-2m-75\le 0$
$\Rightarrow 9m^2-27m+25m-75\le 0$
$\Rightarrow 9m(m-3)+25(m-3)\le 0$
$\Rightarrow (m-3)(9m+25)\le 0$
$\Rightarrow m\in [-\frac{25}{9},3]$

Final values of $m$ for which we have solution is the intersection of two cases
$m\in [-\frac{25}{9},3]\cap (-\infty ,-3)\cup (3,\infty )$
$m\in \varphi$