The number of integral values of x satisfying | vedclass

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Question

(B) We have two inequalities to solve: $5x - 1 < (x + 1)^2$ and $(x + 1)^2 < 7x - 3$. For the first inequality: $5x - 1 < x^2 + 2x + 1$ $x^2 - 3x + 2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We have two inequalities to solve: $5x - 1 For the first inequality: $5x - 1 $x^2 - 3x + 2 > 0$ $(x - 1)(x - 2) > 0$ This implies $x \in (-\infty, 1) \cup (2, \infty)$. For the second inequality: $(x + 1)^2 $x^2 + 2x + 1 $x^2 - 5x + 4 $(x - 1)(x - 4) This implies $x \in (1, 4)$. Taking the intersection of both intervals: $x \in ((-\infty, 1) \cup (2, \infty)) \cap (1, 4) = (2, 4)$. The integers in the interval $(2, 4)$ is only $x = 3$. Thus,there is only $1$ integral value.

The number of integral values of $x$ satisfying $5x - 1 < (x + 1)^2 < 7x - 3$ is

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