Solving Equations Involving Square Roots and Exponents: A Step-by-Step Guide

Equations involving square roots and exponents can often be complex, but with a systematic approach, they can be solved efficiently. Let's explore a detailed example and the techniques used to solve such equations.

Introduction

This article aims to guide you through the process of solving an equation that involves both square roots and exponents. We will derive a solution to the equation y 2 3 x , and we will transform it into a more manageable form.

Step-by-Step Solution

First, let's rewrite the equation y 2 3 x as follows:

Express it as y frac whiteColor{1}{y} 4 . Multiply both sides by y to get y 2 -4y10 . Next, we solve this quadratic equation y 2 -4y10 .

Now, let's solve the equation in detail:

Transformation and Further Simplification

Given the equation y frac whiteColor{1}{y} 4 , multiply by y to get:

y frac whiteColor{1}{y} y 4 y

Simplifying, we get:

y 2 -4y10

Solving the Quadratic Equation

The quadratic equation can be solved using the quadratic formula:

y 2 -4y10 plusmn; 4 pmsqrt{16-4} x 2 plusmn; 4 pmsqrt{12} x x plusmn; 4 pmsqrt{12} x x

Simplifying further, we find:

y 2 -4y10 2pmsqrt{3}

This gives us two potential solutions for y , namely 2pmsqrt{3} .

Backward Substitution and Finding x

Now, we substitute 2pmsqrt{3} back into the original equation:

1) For y 2 sqrt{3} , we find:

y 2 sqrt{3} x 1

2) For y 2 -sqrt{3} , we get:

y 2 -sqrt{3} x -1

Therefore, the potential solutions for x are xpm 1 .

Conclusion

Solving complex equations involving square roots and exponents requires a systematic approach. By transforming the original equation and solving the resulting quadratic equation, we were able to find the values of x . The solutions to the given problem are xpm 1 .

References and Further Reading

For more detailed insights and practice, consider exploring additional resources on solving equations involving square roots and exponents.