Problem 14 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

What is the value of \(x\) satisfying the equation \(16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x} ?\) (a) \(a / 2\) (b) \(a / 3\) (c) \(a / 4\) (d) 0

Step-by-Step Solution

Answer

The answer is \( x = \frac{a}{3} \), which corresponds to option (b).

Step 1: Set up the initial problem

We start with the equation given in the problem: \[ 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x} \]. Our goal is to find the value of \(x\) that satisfies this equation.

Step 2: Introduce a substitution

To simplify the problem, let's substitute \( y = \frac{a-x}{a+x} \). This gives us two equivalent equations: \( y^3 = \frac{a+x}{16(a-x)} \) and \( y = \frac{a-x}{a+x} \).

Step 3: Express \(y^3\) in terms of \(y\)

Using both equations, we equate \(y^3\) and simplify: \[ y^3 = \frac{1}{16y} \].

Step 4: Solve the equation \(y^4 = \frac{1}{16}\)

Multiply both sides by \(y\), resulting in \( y^4 = \frac{1}{16} \). Solve for \(y\) by taking the fourth root of both sides: \( y = \frac{1}{2} \).

Step 5: Substitute back to find \(x\)

Having \( y = \frac{1}{2} \), substitute back into \( \frac{a-x}{a+x} = \frac{1}{2} \). This results in \( 2(a-x) = a+x \), leading to \( 2a - 2x = a + x \).

Step 6: Solve for \(x\)

Rearrange the equation from the previous step: \( 2a - a = x + 2x \), which simplifies to \( a = 3x \). Thus, \( x = \frac{a}{3} \).

Key Concepts

Substitution Method

The substitution method is an effective technique used to simplify complex equations by introducing new variables. In algebra, substitution allows us to handle equations with intricate expressions or multiple terms more comfortably.

For example, in this case, dealing with the equation \[ 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x} \]can be quite overwhelming. However, by setting a new variable like \( y = \frac{a-x}{a+x} \), we streamline the process.

• This reduces the complexity, transforming the original problem into equations involving \( y \) instead of fractions.
• The problem becomes more manageable, focusing on simplification before tackling the main solution.

Substitution is especially valuable in scenarios where solving directly would require intensive calculations. It’s a powerful tool to develop an understanding of the broader problem before diving into detailed solutions.

Simplifying Equations

Simplifying equations is an essential skill in algebra that helps bring clarity to problems by reducing expressions to their simplest form.

Take, for instance, transforming equations involving powers and fractions into a more straightforward form, such as turning \( y^3 = \frac{1}{16y} \) into \( y^4 = \frac{1}{16} \).

• This type of simplification often involves combining terms, minimizing fractions, or re-expressing powers to find a core solution.
• It helps identify relationships between variables, making it possible to solve for one variable in terms of the others.

By simplifying equations, we strip away complexities, thus gaining a clearer path to solving the problem. The aim is to make equations easier to handle, paving the way for more intuitive solutions.

Solving Rational Equations

Rational equations involve fractions that have polynomials in their numerators and denominators. Solving these equations requires careful handling to avoid errors.

One key aspect of solving rational equations is making denominators consistent or eliminating them where possible. In our solution, by working with \( \frac{a-x}{a+x} \)and its inverse, we were able to find shared terms and balance them effectively.

• It often involves multiplying across to get rid of denominators or using fractions to simplify equations.
• Careful steps ensure that equivalent expressions maintain their mutual relationships, and one-to-one relations help isolate terms for results.

It’s crucial to check your solutions back in the original equation to ensure that no extraneous solutions appear due to mathematical manipulations. Mastering rational equations equips you with the skills to tackle many algebraic problems effectively.