The number of integral solutions of $2(x^2+\frac{1}{x^2})-7(x+\frac{1}{x})+9=0$ when $x\ne 0$  is

$2(x^2+\frac{1}{x^2})-7(x+\frac{1}{x})+9=0\Rightarrow 2[{(x+\frac{1}{x})}^2-2]-7(x+\frac{1}{x})+9=0$

$\Rightarrow 2(y^2-2)-7y+9=0$  where   $y=x+\frac{1}{x}\Rightarrow 2y^2-7y+5=0\Rightarrow (y-1)(2y-5)=0$

$(x+\frac{1}{x}-1)(2x+\frac{2}{x}-5)=0\Rightarrow (x^2-x+1)(2x^2-5x+2)=0$

$\Rightarrow x=\frac{1\pm \sqrt{1+4}}{2}or(x-2)(2x-1)=0\Rightarrow x=\frac{1\pm \sqrt{3}i}{2}orx=2or1/2$

 $\therefore$ The number of integral solutions =1.