The number of positive integral solutions of x^2+9<(x+3)^2 < 8x + 25,…

We have,

$x^2+9<(x+3)^2 < 8x + 25$

$⇒x^2+9<x^2+6x+9 <8x + 25$

$⇒x^2+9<x^2+6x+9$ and $x^2+6x+9 <8x+25$

$⇒6x>0$ and $x^2 -2x-16 <0$

$x^2-2x-16=0$

so $D=\sqrt{2^2+4×16}=2\sqrt{17}$

$x = \frac{2±2\sqrt{17}}{2}=1+\sqrt{17}$

for $(x^2-2x-16)<0$

$⇒ 1-\sqrt{17}<x<1+\sqrt{17}$

$⇒x>0$ and $1-\sqrt{17} <x<1+ \sqrt{17}$

$⇒0<x<1+ \sqrt{17}⇒x=1,2,3,4,5$