Animated Solution for Mathematics - Quadratic Equations: Find all integers $x$ for which $(5x-1)<(x+1{)}^2<(7x-3)$.

Welcome, future engineer. Today, we are not just solving an inequality; we are embarking on a journey of logical precision. The problem before us is a classic JEE Advanced challenge: $(5x-1)<(x+1{)}^2<(7x-3)$.

At first glance, it looks like a simple sandwich of expressions, but it is a trap for the careless. Let us dissect it with the precision of a surgeon.

The most common mistake students make is trying to manipulate all three parts of the inequality simultaneously. You cannot do that here because the variable $x$ is squared.

Instead, we must treat this as a system of two simultaneous constraints. We need to find the values of $x$ that satisfy both:

1. $5x-1<(x+1{)}^2$

2. $(x+1{)}^2<7x-3$

Only the intersection of these two solution sets will give us the final answer.

Let us expand the middle term first. Using the identity $(a+b{)}^2=a^2+2ab+b^2$, we transform $(x+1{)}^2$ into $x^2+2x+1$.

Now, look at our first inequality:

To solve this, we move everything to one side to form a standard quadratic inequality:

This simplifies to:

Factoring this, we get $(x-1)(x-2)>0$. Using the Wavy Curve Method, we know that for the product to be positive, $x$ must lie outside the roots. Thus, our first solution set is:

Now, we turn our attention to the second constraint: $x^2+2x+1<7x-3$. Again, we bring all terms to one side:

This simplifies to:

Factoring this, we get $(x-1)(x-4)<0$. Here, the inequality is less than zero, which means $x$ must lie between the roots. Our second solution set is:

This is the moment of truth. We have two conditions: $x$ must be in $(-\infty ,1)\cup (2,\infty )$ AND $x$ must be in $(1,4)$.

Visualize the number line. The first set excludes the region between $1$ and $2$. The second set is strictly between $1$ and $4$.

Where do they overlap? The only region that satisfies both is the interval:

The question asks for all integers $x$ in this interval. Looking at the open interval $(2,4)$, the only integer contained within is $3$.

Let us verify this. If $x=3$, then:

Indeed, $14<16<18$. The logic holds perfectly. You have successfully navigated the constraints and found the truth hidden in the algebra.