Solving the Inequality racΒ²(2x – 1)(x + 3) > 1: A Step-by-Step Guide

When faced with the inequality racΒ²(2x – 1)(x + 3) > 1, many students and math enthusiasts wonder how to approach it efficiently. This article walks you through solving the inequality racΒ²(2x – 1)(x + 3) > 1 step-by-step, including key concepts and common pitfalls to avoid.

Understanding the Context

Understanding the Inequality

The inequality to solve is:
racΒ²(2x – 1)(x + 3) > 1

Here, racΒ²(2x – 1)(x + 3) means [ (2x – 1)(x + 3) ]Β², the square of the expression (2x – 1)(x + 3). This quadratic expression is inside a square, making it non-negative regardless of the signs of the factors. The inequality compares this squared expression to 1, so we're essentially solving when a squared term exceeds 1.

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Image Gallery

Solved integral x + 1 / x^3 - 2x^2 + x dx x (x^2 - 2x +1) x | Chegg.com
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Key Insights

Step 1: Rewrite the Inequality Clearly

Start by clearly writing the inequality in standard form:
(2x – 1)(x + 3)Β² > 1

This step makes it easier to analyze the behavior of the expression.

Step 2: Move All Terms to One Side

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Final Thoughts

To prepare solving, bring 1 to the left side:
(2x – 1)(x + 3)Β² – 1 > 0

Now we want to solve when this expression is greater than zero.

Step 3: Analyze the Function as a Combined Function

Let:
f(x) = (2x – 1)(x + 3)Β² – 1

Our goal: solve f(x) > 0.

First, expand (x + 3)Β²:
==> (x + 3)Β² = xΒ² + 6x + 9

Now substitute:
f(x) = (2x – 1)(xΒ² + 6x + 9) – 1

Multiply out:
f(x) = (2x – 1)(xΒ² + 6x + 9) – 1
= 2x(xΒ² + 6x + 9) – 1(xΒ² + 6x + 9) – 1
= 2xΒ³ + 12xΒ² + 18x – xΒ² – 6x – 9 – 1
= 2xΒ³ + 11xΒ² + 12x – 10

So the inequality becomes:
2xΒ³ + 11xΒ² + 12x – 10 > 0