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Chapter 11: Problem 35

Solve. $$ y^{1 / 3}-y^{1 / 6}-6=0 $$

Short Answer

Expert verified
The solution is \( y = 729 \).

Step by step solution

01

- Substitute Variables

Let us introduce a substitution to make the equation easier to solve. We set: \[ u = y^{1/6} \]. This implies that \[ u^2 = y^{1/3} \].
02

- Rewrite the Equation

Using the substitution from Step 1, we rewrite the equation: \[ u^2 - u - 6 = 0 \].
03

- Solve the Quadratic Equation

To solve the quadratic equation, we factorize it: \[ u^2 - u - 6 = (u - 3)(u + 2) = 0 \]. This gives us the solutions for \( u \): \[ u - 3 = 0 \implies u = 3 \] and \[ u + 2 = 0 \implies u = -2 \].
04

- Back-Substitute and Solve for y

Recall that \( u = y^{1/6} \). We now solve for \( y \) with the values of \( u \): 1. For \( u = 3 \): \[ 3 = y^{1/6} \] \[ y = 3^6 = 729 \] 2. For \( u = -2 \): \[ -2 = y^{1/6} \] Since a negative number can't be the 6th root of a real number, we discard \( u = -2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a great way to simplify complex equations by replacing a part of the equation with a new variable. In this exercise, we simplify the equation \( y^{1 / 3}-y^{1 / 6}-6=0 \) by setting \( u = y^{1/6} \). This results in the easier equation \( u^2 - u - 6 \). Substitution helps to convert complicated equations into more manageable forms.
  • Replace complicated expressions with a single variable.
  • Make sure to revert to the original variable after solving the simpler equation.
  • Always verify and back-substitute to ensure the solution satisfies the original equation.

This method helps in avoiding errors and makes the problem-solving process much clearer.
Quadratic Equation
A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \). In this problem, we encounter the quadratic equation \( u^2 - u - 6 = 0 \) after substitution. Quadratics are solved using methods like factoring, completing the square, or the quadratic formula. In this case, we use factoring because it appears to be the simplest approach.
  • The general form: \( ax^2 + bx + c = 0 \).
  • Solutions usually involve finding values of \( x \) that satisfy the equation.
  • Quadratic equations can have two, one, or no real solutions.

Here, by factoring \( (u - 3)(u + 2) = 0 \), we find the solutions to be \( u = 3 \) and \( u = -2 \). However, further steps reveal that only \( u = 3 \) is valid when back-substituting.
Factoring
Factoring is a method used to solve quadratic equations by expressing the quadratic expression as a product of its factors. In this exercise, we factorize \( u^2 - u - 6 \) as \( (u - 3)(u + 2) \). This technique transforms the equation into a form that's easy to solve by finding the roots.
  • Identify factors of \( c \) (constant term) that add up to \( b \) (the coefficient of \( x \)).
  • Expression of the quadratic equation as products of binomials.
  • Set each binomial equal to zero to solve for the variable.

Factoring is especially useful for simple quadratic equations where the roots are integers. After factoring \( (u - 3)(u + 2) = 0 \), we get the solutions \( u = 3 \) and \( u = -2 \). But as we solve for \( y \), only \( u = 3 \) gives a valid solution, \( y = 729 \).

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