Problem 37 - Algebra for College Students

Question

Solve each equation. $$3 x^{4}+17 x^{2}+20=0$$

Step-by-Step Solution

Answer

No real solutions.

Step 1: Substitution

Let us make a substitution to simplify the equation. Set \( y = x^2 \). Therefore, the equation \( 3x^4 + 17x^2 + 20 = 0 \) becomes \( 3y^2 + 17y + 20 = 0 \).

Step 2: Solve the Quadratic Equation

We now have a quadratic equation \( 3y^2 + 17y + 20 = 0 \). We will use the quadratic formula to solve it, where \( a = 3 \), \( b = 17 \), and \( c = 20 \). The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 3: Compute the Discriminant

Calculate the discriminant \( b^2 - 4ac \) using the values \( b = 17 \), \( a = 3 \), and \( c = 20 \): \[ b^2 - 4ac = 17^2 - 4\cdot3\cdot20 = 289 - 240 = 49 \].

Step 4: Apply the Quadratic Formula

Since the discriminant is 49, compute the solutions for \( y \):\[ y = \frac{-17 \pm \sqrt{49}}{6} = \frac{-17 \pm 7}{6} \].This gives two solutions: \( y_1 = \frac{-17 + 7}{6} = \frac{-10}{6} = -\frac{5}{3} \)and \( y_2 = \frac{-17 - 7}{6} = \frac{-24}{6} = -4 \).

Step 5: Back Substitute to Original Variable

Since \( y = x^2 \), we now substitute back to find \( x \):- For \( y_1 = -\frac{5}{3} \), there is no real solution since the square of a real number cannot be negative.- For \( y_2 = -4 \), similarly, no real solution exists. Thus, there are no real solutions in \( x \).

Key Concepts

Quadratic Equation

A quadratic equation is an algebraic expression of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The key feature of a quadratic equation is the \( x^2 \) term. This term makes the graph of the equation a parabola, which is a u-shaped curve. Quadratic equations are central in algebra and appear in various mathematical and real-world contexts.
To solve a quadratic equation, you can use methods such as:

• Factoring: Expressing \( ax^2 + bx + c \) as a product of two binomials, if possible.
• Using the Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the Square: Rearranging the equation to create a perfect square trinomial.

The quadratic formula is particularly useful when factoring is complicated or not possible. The solutions to the quadratic equation are known as the "roots" or "zeros" of the equation.

Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is found inside the square root part of the quadratic formula: \( b^2 - 4ac \). The discriminant helps determine the nature of the roots of the equation without actually solving it entirely, as it tells you how many and what kind of solutions you have.
Here's how to interpret the discriminant:\[\begin{align*}1. &\ b^2 - 4ac > 0: \text{Two distinct real solutions.}\2. &\ b^2 - 4ac = 0: \text{One real solution (or a repeated real root).}\3. &\ b^2 - 4ac < 0: \text{No real solutions (solutions are complex numbers).}\\end{align*}\]In our example, the discriminant was calculated as 49, which is greater than zero. This indicates that the substituted quadratic equation in terms of \( y \) has two distinct solutions.

Substitution Method

The substitution method is a helpful technique for solving complex equations by simplifying them into a more manageable form. In the context of this problem, substitution was used to turn a quartic equation, \( 3x^4 + 17x^2 + 20 = 0 \), into a quadratic equation by setting \( y = x^2 \).
This process involves:

• Choosing a substitution that simplifies the expression. Here, \( x^2 \) was replaced with \( y \), reducing the degree of the polynomial from four to two.
• Solving the resulting simpler equation. In our case, the equation \( 3y^2 + 17y + 20 = 0 \) was solved using the quadratic formula.
• Reversing the substitution to find the solution in terms of the original variable. After finding \( y \), substituting back to find \( x \) was done, revealing no real solutions.

This method is particularly effective when dealing with polynomials that can be transformed into a recognizable form like a quadratic equation.

Real Solutions

Real solutions refer to the values of \( x \) that satisfy the equation without involving complex numbers. In terms of quadratic equations, a solution is considered real when it does not include the imaginary unit \( i \), which is defined as the square root of \(-1\).
For a quadratic equation, real solutions occur when the discriminant \( b^2 - 4ac \) is non-negative:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution.
• If \( b^2 - 4ac < 0 \), there are no real solutions, implying the solutions are complex.

In the particular exercise provided, the computed discriminant permitted solutions in terms of \( y \). However, back-substitution revealed no real values for \( x \) because the scenarios resulted in seeking the square root of a negative number, which does not exist among real numbers. Understanding real versus complex solutions is crucial in algebra as it can impact the implications and applications of a mathematical problem in real-world contexts.