Find all integral values of x for which $(5 x - 1) < {(x + 1)}^{2} < (7 x - 3) .$

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Solution

from $(5 x - 1) < {(x + 1)}^{2}$
$\Rightarrow x^{2} + 2 x + 1 > 5 x - 1$
$\Rightarrow x^{2} - 3 x + 2 > 0$
$\Rightarrow (x - 1) (x - 2) > 0$ ...(1)
from ${(x + 1)}^{2} < (7 x - 3)$
$\Rightarrow x^{2} - 5 x + 4 > 0$
$\Rightarrow (x - 1) (x - 4) > 0$ ...(2)
$x < 1$ or $x > 2$ and $1 < x < 4$
$∴ 2 < x < 4$
Thus, integral value is $x = 3$

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