Problem 52 - Algebra and Trigonometry

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. $$x^{\frac{2}{5}}+x^{\frac{1}{5}}-6=0$$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2/5} + x^{1/5} - 6 = 0\) are \(x = 32\) and \(x = -243\)

Step 1: Identify the substitution

In the given equation, \( x^{2/5} + x^{1/5} - 6 = 0 \), it is observed that one term is a square of another. Therefore, taking \( y = x^{1/5} \) will simplify the equation.

Step 2: Apply the substitution to the equation

By substituting \( y = x^{1/5} \) into the original equation, it transforms into \( y^2 + y - 6 = 0\), which is a quadratic equation.

Step 3: Solve the quadratic equation

The quadratic equation \( y^2 + y - 6 = 0 \) can be factored into \( (y - 2)(y + 3) = 0 \). Set each factor equal to zero and solve for y. This gives two solutions: \( y1 = 2 \) and \( y2 = -3 \).

Step 4: Substitute y back into x

After finding the values of y, substitute them back into \( y = x^{1/5} \) to get the values of x. Solving it gives \( x1 = y1^5 = 32 \) and \( x2 = y2^5 = -243 \).

Step 5: Check the solutions

To validate the solutions, substitute them back into the original equation \( x^{2/5} + x^{1/5} - 6 = 0 \). After substituting \( x1 = 32 \) and \( x2 = -243 \) into the equation, it verifies that both satisfy the original equation.

Key Concepts

Quadratic Equations

Quadratic equations are an essential part of algebra. They follow the standard form of \( ax^2 + bx + c = 0 \). Here, \( a, b, \) and \( c \) are constants. The most interesting thing about quadratic equations is that they can often be solved in more than one way.

Quadratic equations can have:

• Two real solutions
• One real solution
• Two complex solutions

Graphs of these equations typically form a parabola. This means the equation can open upwards or downwards, depending on the sign of \( a \).
Understanding the shape and solutions of quadratics is crucial for solving real-world problems, from ballistics to economics.

Factoring Quadratics

Factoring is one of the main techniques to solve quadratic equations. It involves expressing the quadratic as a product of two binomials. For instance, if you have an equation in the form of \( y^2 + y - 6 = 0 \), you can factor it as \( (y - 2)(y + 3) = 0 \).

The advantage of factoring is that it allows you to quickly find the values of the variable that make the equation true:

• Identify simpler expressions such as \( y - 2 \) and \( y + 3 \).
• Set each of these expressions equal to zero.
• Solve for the variable \( y \) to find the solutions.

Factoring is not always possible, especially if the quadratic does not factor into integers easily. In such cases, other methods like using the quadratic formula or completing the square can be more useful.

Equation Solving

Equation solving is the process of finding the values of variables that satisfy a given equation. It requires a good grasp of different algebraic techniques and a systematic approach.

Here’s a quick guide to solving equations:

• Isolate the variable. Start by manipulating the equation to get the variable on one side.
• Apply substitution if complicated terms are involving powers, like in the case \( x^{1/5} \).
• Verify your solutions by substituting them back into the original equation to ensure they work.

Understanding these steps is key to solving not just algebraic equations but, indeed, any math or science problems that come your way.

Exponent Rules

Understanding exponent rules is foundational for algebraic substitution and solving equations like \( x^{2/5} + x^{1/5} - 6 = 0 \). These rules help manage powers of variables efficiently.

Key exponent rules include:

• Product of powers: \( a^m \times a^n = a^{m+n} \)
• Power of a power: \( (a^m)^n = a^{m \times n} \)
• Power of a product: \( (ab)^n = a^n b^n \)
• Negative exponents: \( a^{-m} = \frac{1}{a^m} \)
• Fractional exponents: \( a^{m/n} \) represents the \( n \)-th root of \( a^m \)

These rules make it easier to manipulate and simplify expressions involving exponents, which is crucial when trying to solve more complex algebraic equations.