Question

The number of positive integral solutions of $x^{2}+9 < (x+3)^{2} < 8 x+25$, is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

We have, $ {{x}^{2}}+9 < {{(x+3)}^{2}} < 8x+25 $
$ \Rightarrow $ $ {{x}^{2}}+9 < {{x}^{2}}+6x+9 < 8x+25 $
$ \Rightarrow $ $ {{x}^{2}}+9<{{x}^{2}}+6x+9 $
and $ {{x}^{2}}+6x+9 < 8x+25 $
$ \Rightarrow $ $ 6x > 0 $ and $ {{x}^{2}}-2x-16 < 0 $
$ \Rightarrow $ $ x > 0 $ and $ 1-\sqrt{17} < x < 1 +\sqrt{17} $
$ \Rightarrow $ $ 0 < x < 1+\sqrt{17} $
$ \therefore x = 1, 2, 3, 4, 5$