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Magazine Chapter 3: Problem 53
Solve the polynomial inequality and state your answer using interval notation. \(2 x^{4}>5 x^{2}+3\)
Short Answer
Expert verified
The solution is \((-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)\).
Step by step solution
01
Move All Terms to One Side
Start by moving all terms to one side of the inequality. We have: \[ 2x^4 > 5x^2 + 3 \]Subtract \(5x^2\) and \(3\) from both sides:\[ 2x^4 - 5x^2 - 3 > 0 \].
02
Consider the Corresponding Equation
Solve the equation \(2x^4 - 5x^2 - 3 = 0\) to find the critical points. These will help us determine the intervals to check the sign of the inequality.
03
Set a Substitution
Let \(y = x^2\). Then the equation becomes:\[ 2y^2 - 5y - 3 = 0 \].
04
Solve the Quadratic Equation
Use the quadratic formula to solve \(2y^2 - 5y - 3 = 0\):\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 2\), \(b = -5\), and \(c = -3\).
05
Apply the Quadratic Formula
Calculate:\[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 2 \times (-3)}}{2 \times 2} \]\[ y = \frac{5 \pm \sqrt{25 + 24}}{4} \]\[ y = \frac{5 \pm \sqrt{49}}{4} \]\[ y = \frac{5 \pm 7}{4} \].
06
Determine the Roots of the Quadratic
The roots are:\[ y = \frac{12}{4} = 3 \]\[ y = \frac{-2}{4} = -0.5 \].
07
Find the Values of x
Since \(y = x^2\), solve \(x^2 = 3\) and \(x^2 = -0.5\). For \(x^2 = 3\), \(x = \pm \sqrt{3}\). As \(x^2 = -0.5\) has no real solutions, consider only \(\pm \sqrt{3}\).
08
Test Intervals Based on Critical Points
The critical points are \(-\sqrt{3}\) and \(\sqrt{3}\). Test intervals: 1. \((-\infty, -\sqrt{3})\)2. \((-\sqrt{3}, \sqrt{3})\)3. \((\sqrt{3}, \infty)\).
09
Choose Test Points
Select test points from each interval, such as:1. \(-2\) from \((-\infty, -\sqrt{3})\)2. \(0\) from \((-\sqrt{3}, \sqrt{3})\)3. \(2\) from \((\sqrt{3}, \infty)\).
10
Evaluate the Inequality at Test Points
Substitute each test point into \(2x^4 - 5x^2 - 3\):- For \(x = -2\): \(2(-2)^4 - 5(-2)^2 - 3 = 32 - 20 - 3 = 9 > 0\)- For \(x = 0\): \(2(0)^4 - 5(0)^2 - 3 = -3 < 0\)- For \(x = 2\): \(2(2)^4 - 5(2)^2 - 3 = 32 - 20 - 3 = 9 > 0\).
11
Determine the Solution Intervals
The inequality is satisfied in intervals where the test point evaluation is positive. Thus, the solution in interval notation is:\((-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
To solve quadratic equations like \(2y^2 - 5y - 3 = 0\), we often use the quadratic formula. It is given by:
By calculating, we determine that the solutions (or roots) for \(y\) are \(3\) and \(-0.5\). These solutions are critical as they help in identifying intervals to test in the polynomial inequality.
- \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
By calculating, we determine that the solutions (or roots) for \(y\) are \(3\) and \(-0.5\). These solutions are critical as they help in identifying intervals to test in the polynomial inequality.
Critical Points
Critical points are the values of \(x\) that zero out a function or change its direction. They are important for determining where the polynomial inequality might change sign.
In our example, we obtained \(y\) values \(3\) and \(-0.5\) by substituting \(x^2 = 3\) and noting that \(x^2 = -0.5\) has no real roots since a real number squared cannot be negative. Solving \(x^2 = 3\) gives us critical points \(x = \pm \sqrt{3}\), which help define the test intervals for the inequality.
In our example, we obtained \(y\) values \(3\) and \(-0.5\) by substituting \(x^2 = 3\) and noting that \(x^2 = -0.5\) has no real roots since a real number squared cannot be negative. Solving \(x^2 = 3\) gives us critical points \(x = \pm \sqrt{3}\), which help define the test intervals for the inequality.
Interval Notation
Interval notation is a method to describe a range of values in a compact form. It uses brackets to express whether endpoints are included or excluded:
The solution is written as \(( -\infty, -\sqrt{3} ) \cup ( \sqrt{3}, \infty )\), combining two open intervals, showing all \(x\) values where the inequality holds.
- "(" and ")" mean endpoints are not included (open interval).
- "[" and "]" indicate endpoints are included (closed interval).
The solution is written as \(( -\infty, -\sqrt{3} ) \cup ( \sqrt{3}, \infty )\), combining two open intervals, showing all \(x\) values where the inequality holds.
Test Intervals
Test intervals come from the critical points of a polynomial. They define ranges where we need to test whether the inequality is positive or negative.
For \(x = -\sqrt{3}\) and \(x = \sqrt{3}\), we divide the real number line into three intervals:
If a test point satisfies the inequality (i.e., leads to a positive result), then the entire interval satisfies the inequality. Thus, solutions are confirmed in the intervals where the function is greater than zero.
For \(x = -\sqrt{3}\) and \(x = \sqrt{3}\), we divide the real number line into three intervals:
- \(( -\infty, -\sqrt{3} )\)
- \(( -\sqrt{3}, \sqrt{3} )\)
- \(( \sqrt{3}, \infty )\)
If a test point satisfies the inequality (i.e., leads to a positive result), then the entire interval satisfies the inequality. Thus, solutions are confirmed in the intervals where the function is greater than zero.
