Problem 39 - Algebra and Trigonometry

Question

\(31-42\) . Solve the equation both algebraically and graphically. $$ 16 x^{4}=625 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{25}{4}\) and \(x = -\frac{25}{4}\).

Step 1: Algebraic manipulation

We start by rewriting the given equation \(16x^4 = 625\). First, divide both sides by 16 to isolate \(x^4\) on one side:\[x^4 = \frac{625}{16}\]

Step 2: Taking the fourth root

Next, solve for \(x\) by taking the fourth root of both sides:\[x = \pm \sqrt[4]{\frac{625}{16}}\] Simplifying, we have:\[x = \pm \frac{\sqrt{625}}{\sqrt{16}}\] Which leads to:\[x = \pm \frac{25}{4}\] Hence, the solutions are \(x = \frac{25}{4}\) and \(x = -\frac{25}{4}\).

Step 3: Graphical solution insight

To solve graphically, consider the functions \(f(x) = 16x^4\) and \(g(x) = 625\). Plot both functions on the same graph. The solutions to \(16x^4 = 625\) are the x-values where the graphs intersect.Both graphs intersect at \(x = \frac{25}{4}\) and \(x = -\frac{25}{4}\), confirming our algebraic solutions.

Step 4: Verification of solutions

Substitute \(x = \frac{25}{4}\) back into the original equation to verify:\[16\left(\frac{25}{4}\right)^4 = 625\] Calculate:\[16 \left(\frac{25}{4}\right)^4 = 16 \times \frac{390625}{256} = 625\]Similarly, substitute \(x = -\frac{25}{4}\) and verify that this is a solution as well. Thus, both solutions checked are correct.

Key Concepts

Fourth Root

Finding the fourth root is a process similar to finding the square root or cube root, but it involves taking the root four times.
In the equation from our exercise, we started by simplifying the power term. We began with \(x^4 = \frac{625}{16}\). The goal here was to isolate \(x\) by reversing the exponentiation.

• A fourth root is notated as \(\sqrt[4]{...}\) and is equivalent to raising a number to the power of \(\frac{1}{4}\).
• In our example, \(\sqrt[4]{625}\) simplifies to 25 because the fourth root of 625 is 5 (since 5 to the power of four is 625).
• Similarly, \(\sqrt[4]{16}\) simplifies to 2, because 2 to the power of four is 16.

Thus, the operation of taking the fourth root of a number provides values which, when raised to the fourth power, return the original number.
In many equations, finding the fourth root is crucial for reducing higher powers and simplifying the problem.

Algebraic Manipulation

Algebraic manipulation is a fundamental process in solving equations. It involves rearranging and simplifying expressions to isolate the variable of interest.
Let's break down how we manipulated the exercise equation algebraically:

• We began with the equation \(16x^4 = 625\). The first step was dividing both sides by 16 to make the expression simpler: \(x^4 = \frac{625}{16}\).
• Once we had the equation in this form, our next step involved solving for \(x\) using roots. We took the fourth root of both sides resulting in: \(x = \pm\sqrt[4]{\frac{625}{16}}\).
• This process allowed us to simplify further by breaking down into square roots: \(x = \pm\frac{\sqrt{625}}{\sqrt{16}} = \pm\frac{25}{4}\).

The power of algebraic manipulation lies in its ability to transform complex equations into simpler, manageable forms, leading to the discovery of the solution.

Graphical Solution

A graphical solution involves using graphs to identify values of \(x\) that satisfy an equation. Here's how this works for our example:

• We considered the functions \(f(x) = 16x^4\) and \(g(x) = 625\). The task was to find where \(f(x)\) and \(g(x)\) have the same value, i.e., the intersection points.
• Graphically, this means plotting both functions on a graph. The intersections occur at the solutions \(x = \frac{25}{4}\) and \(x = -\frac{25}{4}\).
• The benefit of a graphical solution is that it visually confirms the points where the solutions exist and can provide insight into the behavior of the function graph.

Using graphs to solve equations provides a unique, visual perspective which can be especially helpful when handling nonlinear and complex equations.