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The number of positive integral solutions ${x^2} + 9 < {\left( {x + 3} \right)^2} < 8x + 25$ ,is
$
  A.{\text{ 2}} \\
  {\text{B}}{\text{. 3}} \\
  {\text{C}}{\text{. 4}} \\
  {\text{D}}{\text{. 5}} \\
 $

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Last updated date: 16th Jul 2024
Total views: 450k
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Answer
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Hint: In order to solve this question, we will solve the inequalities separately taking one term as constant in order to find out the number of positive integrals.

Complete step-by-step answer:
Given,
${x^2} + 9 < {\left( {x + 3} \right)^2} < 8x + 25$
Taking inequality number (1)
${x^2} + 9 < {\left( {x + 3} \right)^2}$
Or ${x^2} + 9 < {x^2} + 6x + 9$
Or ${x^2} + 9 - {x^2} - 6x - 9 < 0$
Or $ - 6x < 0$
Or $6x > 0$
Or $x > 0 - - - - - - \left( 1 \right)$
Taking inequality number (2)
Or ${x^2} + 6x + 9 < 8x + 25$
Or ${x^2} - 2x - 16 < 0$
Using ,$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a = 1,{\text{ }}b = - 2,{\text{ c = - 16}}$
$x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{2^2} - 4\left( 1 \right)} \left( { - 16} \right)}}{{2\left( 1 \right)}}$
Or $x = \dfrac{{2 \pm \sqrt {4 + 64} }}{2}$
Or $x = \dfrac{2}{2}\left( {1 \pm \sqrt {17} } \right)$
Or $x = 1 \pm \sqrt {17} $
Or $1 - \sqrt {17} < x < 1 + \sqrt {17} $
But it must be a positive integral.
Therefore,
$0 < x < 1 + \sqrt {17} $
So the integer values in our domain are
$1,2,3,4,5$
So, the option $\left( D \right)$ is correct.

Note: Whenever we face these types of questions the key concept is that we have to take the inequalities separately and only the positive values of $x$ is considerable as the number of integrals and in this way we will get our desired answer. Here, in this we did the same thing and thus we got our desired answer.