The variable question - 2

• $x=3$ and $x=-3$ are not solutions

Case 1 $x>3$ $x-3>0\Rightarrow |x-3|=x-3$ $and$ $x^2-9=(x-3)(x+3)\forall x\in Z^+|x-3>0;(x-3)(x+3)>0\Rightarrow x^2-9>0\Rightarrow |x^2-9|=x^2-9$ Putting these values $x-3+\sqrt{x^2-9}-7=0$ $\sqrt{x^2-9}=10-x$ $\cancel{x^2}-9=100\cancel{+x^2}-20x$ $20x=109$ $\boxed{x=\frac{109}{20}}$

Case 2 $x<3$

Case 2 ( $a$ ) $x^2<9$ $|x-3|=-(x-3)=3-x..........(\because x<3 \Rightarrow x-3<0)$ $|x^2-9|=-(x^2-9)=9-x^2..........(\because x^2<9\Rightarrow x^2-9<0)$ Putting these values $3-x+\sqrt{9-x^2}-7=0$ $\sqrt{9-x^2}=4+x$ $9-x^2=x^2+16+8x$ $2x^2+8x+7=0$ As $b^2-4ac=8^2-4(2)(7)=64-56=8>0\therefore$ the equation has real roots

From Vieta's formula Sum of it's roots $\boxed{=\dfrac{-8}{2}=-4}$

Case 2 ( $b$ ) $x^2>9$ $|x-3|=-(x-3)=3-x..........(\because x<3 \Rightarrow x-3<0)$ $|x^2-9|=x^2-9..........(\because x^2>9\Rightarrow x^2-9>0)$ Putting these values $3-x+\sqrt{x^2-9}-7=0$ $\sqrt{x^2-9}=4+x$ $\cancel{x^2}-9=\cancel{x^2}+16+8x$ $0=25+8x$ $\boxed{x=\dfrac{-25}{8}}$

Sum of all real values of $x=-4+\dfrac{109}{20}-{25}{8}=\dfrac{-160+218-125}{40}=\dfrac{218-285}{40}=\dfrac{-67}{40}$ $-67\in Z;40\in Z;\gcd(67,40)=1;67+40=\boxed{107}$

Note that $|x-3| + \sqrt{|x^2-9|} - 7 = \begin{cases} x-3 + \sqrt{x^2 - 9} - 7 = x-10 + \sqrt{x^2-9} & \text{for }x > 3 \\ 3-x + \sqrt{9-x^2} - 7 = -x - 4 + \sqrt{9-x^2} & \text{for }-3 < x \le 3 \\ 3-x + \sqrt{x^2-9} - 7 = -x - 4 + \sqrt{x^2-9} & \text{for } x \le - 3 \end{cases}$

For $x > 3$ :

$\begin{aligned} \sqrt{x^2-9} & = 10 - x & \small \blue{\text{Squaring both sides}} \\ x^2 - 9 & = x^2 - 20x + 100 \\ \implies x & = \frac {109}{20} \end{aligned}$

For $-3<x \le 3$ :

$\begin{aligned} \sqrt{9-x^2} & = x+4 & \small \blue{\text{Squaring both sides}} \\ 9 - x^2 & = x^2 + 8x + 16 \\ 2x^+ 8x + 7 & = 0 \\ \implies x & = \frac {-8 \pm \sqrt{64-56}}4 \\ & = - 2 \pm \frac 1{\sqrt 2} \end{aligned}$

For $x \le -3$ :

$\begin{aligned} \sqrt{x^2-9} & = x+4 & \small \blue{\text{Squaring both sides}} \\ x^2 - 9 & = x^2 + 8x + 16 \\ \implies x & = - \frac {25}8 \end{aligned}$

Therefore the sum of all real roots is $\dfrac {109}{20} - 2 + \dfrac 1{\sqrt 2} - 2 - \dfrac 1{\sqrt 2} - \dfrac {25}8 = - \dfrac {67}{40}$ $\implies m+n = 67+40 = \boxed{107}$ .