Math Problem Statement
x^{2}<\left|4x-3\right|
Solution
We are tasked with solving the inequality:
Step 1: Analyze the absolute value function
The absolute value function, , can be written as a piecewise function:
\begin{cases} 4x - 3 & \text{if } 4x - 3 \geq 0 \quad \text{(i.e., } x \geq \frac{3}{4}) \\ -(4x - 3) & \text{if } 4x - 3 < 0 \quad \text{(i.e., } x < \frac{3}{4}) \end{cases}$$ This breaks the inequality into two cases. ### Step 2: Solve the inequality in two cases #### Case 1: $$x \geq \frac{3}{4}$$ In this case, the inequality becomes: $$x^2 < 4x - 3$$ Rearranging this inequality: $$x^2 - 4x + 3 < 0$$ Factoring the quadratic expression: $$(x - 1)(x - 3) < 0$$ Now, solve the inequality $$(x - 1)(x - 3) < 0$$. This inequality holds when $$1 < x < 3$$. Therefore, for $$x \geq \frac{3}{4}$$, the solution is: $$1 < x < 3$$ #### Case 2: $$x < \frac{3}{4}$$ In this case, the inequality becomes: $$x^2 < -(4x - 3)$$ Simplifying this: $$x^2 + 4x - 3 < 0$$ This quadratic expression can be factored: $$(x + 2)(x - 1) < 0$$ Now, solve the inequality $$(x + 2)(x - 1) < 0$$. This inequality holds when $$-2 < x < 1$$. Since we are considering $$x < \frac{3}{4}$$, the solution for this case is: $$-2 < x < \frac{3}{4}$$ ### Step 3: Combine the two cases From Case 1, we have the solution $$1 < x < 3$$. From Case 2, we have the solution $$-2 < x < \frac{3}{4}$$. Thus, the final solution is the union of these two intervals: $$x \in (-2, \frac{3}{4}) \cup (1, 3)$$ Would you like more details or have any further questions? --- Here are 5 related questions to expand on the problem: 1. How do you solve inequalities involving other absolute value expressions? 2. What is the geometric interpretation of $$x^2$$ versus $$|4x - 3|$$ on a graph? 3. How do you solve inequalities involving quadratic and linear functions without absolute values? 4. Can we apply similar methods to inequalities involving higher degree polynomials? 5. How does the solution change if the inequality is $$x^2 \leq |4x - 3|$$? **Tip:** When solving inequalities involving absolute values, always split the problem into cases based on the definition of the absolute value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Inequalities
Absolute Value
Quadratic Equations
Formulas
Quadratic expression factorization: (x - a)(x - b)
Piecewise definition of absolute value: |x| = x if x ≥ 0; -x if x < 0
Theorems
Properties of inequalities
Factorization of quadratics
Properties of absolute value
Suitable Grade Level
Grades 10-12