Problem 43 - College Algebra Essentials

Question

By making an appropriate substitution. $$ 9 x^{4}=25 x^{2}-16 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = -1\), \(x = 1\), \(x = -4/3\), and \(x = 4/3\).

Step 1: Introduce substitution

Let's start by saying \(y = x^2\). This changes the original equation into \(9y^2 = 25y - 16\).

Step 2: Rewrite the equation in standard form

Rewrite the equation into standard form of a quadratic which is \(ay^2 + by + c = 0\). Hence, it becomes \(9y^2 - 25y + 16 = 0\).

Step 3: Solve for y

Solving the equation \(9y^2 - 25y + 16 = 0\) we obtain \(y = 1\) or \(y = 16/9\). These are the solutions for \(y = x^2\).

Step 4: Substitute y = x^2 into solutions

Substituting \(y = x^2\) into both solutions from the previous step gives us \(x^2 = 1\) and \(x^2 = 16/9\). Solving these, give four solutions for \(x\): \(x = \pm 1\), and \(x = \pm 4/3\).

Key Concepts

Polynomial Equations

Polynomial equations are expressions that involve terms with variables raised to whole number powers. They are one of the most prevalent forms of equations in algebra.
Polynomial equations can range from simple linear equations to more complex forms like quadratic and cubic equations. In this exercise, we are particularly focused on a polynomial equation of degree four, which can be transformed into a quadratic structure through substitution.

• A polynomial equation takes the form: \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0\).
• The degree of the polynomial is determined by the highest power of the variable, which features prominently in the equation.
• Our exercise involves transforming a polynomial equation of degree four: \(9x^4 - 25x^2 + 16 = 0\).

Understanding polynomial equations prepares us to use techniques like substitution and factorization effectively to find solutions.

Substitution Method

The substitution method is a useful algebraic technique that simplifies equations, making them easier to solve. Essentially, this method involves replacing a part of the equation with a new variable to reduce its complexity.
In this exercise:

• We substitute \(y = x^2\). This reduces the original problem \(9x^4=25x^2-16\) into a simpler quadratic form \(9y^2 = 25y - 16\).
• Substitution transforms the polynomial equation to a more manageable format, enabling us to use methods appropriate for quadratic equations.
• Once solved, the new variable, \(y\), is substituted back to find the values of the original variable, \(x\).

The substitution method is valuable because it enables solving equations that initially appear too complex or have higher degrees.

Quadratic Equations

Quadratic equations, given in the standard form \(ay^2 + by + c = 0\), are a central concept in algebra. They are characterized by the highest power of the variable being two. In our exercise, we transformed a quartic equation, using substitution, into a quadratic: \(9y^2 - 25y + 16 = 0\).
Quadratic equations can be solved using various methods:

• Factoring: Expressing the quadratic as a product of two binomials.
• Quadratic Formula: Finding solutions using \(y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
• Completing the Square: Rewriting the equation in the form \((y-h)^2 = k\) and solving for \(y\).

For our specific equation, applying one of the above methods yields two solutions for \(y\), which we then use to find solutions for \(x\). Understanding quadratic equations is essential for handling more advanced algebraic problems.

Solving Equations

Solving equations involves finding the values of the unknown variables that satisfy the equation. It requires a comprehensive understanding of various methods and techniques.
In the given exercise, solving combines a series of methods:

• First, use substitution to simplify the equation to a form that's easier to manage.
• Next, employ techniques suitable for quadratic equations to find possible solutions for \(y\).
• Finally, substitute back the variable from the earlier step to reveal the possible values of \(x\), giving us \(x = \pm 1\) and \(x = \pm \frac{4}{3}\).

Understanding how to solve complex equations step by step equips students with the ability to tackle diverse algebraic problems confidently. Real proficiency comes through practice and mastering these concepts.