Problem 40 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$4 x^{4}-33 x^{2}+50=0$$

Step-by-Step Solution

Answer

Solutions are \(x = \{-2.5, -\sqrt{2}, \sqrt{2}, 2.5\}\) as individual points.

Step 1: Substitute Variable

Let \( u = x^2 \). Then, rewrite the equation as \( 4u^2 - 33u + 50 = 0 \). This substitution simplifies the quartic equation to a quadratic one.

Step 2: Identify Quadratic Form

We now have a quadratic equation in the form \( 4u^2 - 33u + 50 = 0 \). The objective is to solve for \( u \) using the quadratic formula.

Step 3: Apply the Quadratic Formula

The quadratic formula is \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 4 \), \( b = -33 \), and \( c = 50 \). Substitute these values to solve for \( u \).

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \). For \( 4u^2 - 33u + 50 \), the discriminant is \((-33)^2 - 4 \cdot 4 \cdot 50 = 1089 - 800 = 289\).

Step 5: Solve for u

With the discriminant calculated, the roots of the quadratic equation are \( u = \frac{33 \pm 17}{8} \). So \( u_1 = \frac{33 + 17}{8} = 6.25 \) and \( u_2 = \frac{33 - 17}{8} = 2 \). Therefore, \( u = 6.25 \) or \( u = 2 \).

Step 6: Re-Substitute and Solve for x

Since \( u = x^2 \), we find the roots for \( x^2 = 6.25 \) and \( x^2 = 2 \). Solving these gives \( x = \pm \sqrt{6.25} = \pm 2.5 \) and \( x = \pm \sqrt{2} \).

Step 7: Express Solutions as Intervals

The solutions for \( x \) are \( x = -2.5, - obreak\sqrt{2}, obreak\sqrt{2}, 2.5 \). In interval notation, these solutions are not continuous, so they can be written as the union of individual points: \( \{-2.5, -\sqrt{2}, \sqrt{2}, 2.5\} \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation in the standard form \( ax^2 + bx + c = 0 \). It is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to calculate the values of \( x \) that make the quadratic equation true. It's applicable to any quadratic equation, as long as you identify the coefficients \( a \), \( b \), and \( c \).To use this formula effectively:

• Ensure the quadratic equation is arranged in standard form \( ax^2 + bx + c = 0 \).
• Identify the coefficients \( a \), \( b \), and \( c \).
• Substitute these values into the formula to find the roots.
• Calculate the discriminant \( b^2 - 4ac \) as part of the formula process.

Discriminant

The discriminant in the quadratic formula is given by \( b^2 - 4ac \) and plays a crucial role in determining the nature of the roots of the quadratic equation.Here's what the discriminant tells us:

• If \( b^2 - 4ac > 0 \): There are two distinct real numbers as the roots. The parabola intersects the x-axis at two points.
• If \( b^2 - 4ac = 0 \): There is exactly one real root, sometimes called a double root. The parabola just touches the x-axis at one point.
• If \( b^2 - 4ac < 0 \): There are no real roots, as the parabola does not intersect the x-axis.

In our example, the discriminant \( 289 \) shows that there are two distinct real roots for the equation.

Interval Notation

Interval notation is a way of describing a set of numbers along a continuous interval. It is often used to express solutions to inequalities or the domain and range of a function.For example:

• \((a, b)\) represents all numbers between \(a\) and \(b\), not including \(a\) and \(b\).
• \([a, b]\) includes \(a\) and \(b\), along with all numbers in between.
• The union of separate intervals can be expressed using the union symbol \(\cup\).

In our context, the solutions \(-2.5, -\sqrt{2}, \sqrt{2}, 2.5\) are not part of a continuous interval. Therefore, we describe them as individual points using set notation: \(\{-2.5, -\sqrt{2}, \sqrt{2}, 2.5\}\). This approach captures the idea that these solutions don't lie within a single, uninterrupted range.

Polynomial Equations

Polynomial equations are algebraic expressions set up as equations, involving variables and coefficients, where the variable's powers are whole numbers. The most common ones, like quadratic equations, are of the form \( ax^n + bx^{n-1} + ... + k = 0 \).Characteristics of polynomial equations include:

• Degree: Determined by the highest power of the variable, like a fourth-degree equation is quartic.
• Simplicity: Allows rearranging complex equations into simpler forms, often reducing the degree of the polynomial temporarily, like substituting \( u = x^2 \) to simplify \( 4x^4 - 33x^2 + 50 \) to a quadratic equation.
• Solving: Techniques vary based on degree—from factoring and the quadratic formula to numerical methods for higher-degree equations.

In solving the given quartic equation, we used a common strategy of simplifying the equation through substitution, making it more manageable by leveraging what we know about quadratics.