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Chapter 8: Problem 80

The following exercises are not grouped by type. Solve each equation. $$x^{6}-10 x^{3}=-9$$

Short Answer

Expert verified
The solutions are \( x = 1 \) and \( x = \root{3}{9} \).

Step by step solution

01

Substitute

Let's start by substituting a new variable to simplify the equation. Set \( y = x^3 \). This transforms the equation into \( y^2 - 10y = -9 \).
02

Rewrite the Equation

Rewrite the equation from Step 1 into standard quadratic form: \( y^2 - 10y + 9 = 0 \).
03

Factor the Quadratic Equation

Factor the equation \( y^2 - 10y + 9 = 0 \). The equation factors to \( (y - 1)(y - 9) = 0 \).
04

Solve for y

Solve the factored equation: \( y - 1 = 0 \) or \( y - 9 = 0 \). Thus, \( y = 1 \) or \( y = 9 \).
05

Back-Substitute

Recall that \( y = x^3 \). Therefore, \( x^3 = 1 \) or \( x^3 = 9 \).
06

Solve for x

Solve the equations for \( x \): \( x^3 = 1 \) gives \( x = 1 \), and \( x^3 = 9 \) gives \( x = \root{3}{9} \).

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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 powerful tool for solving polynomial equations. It involves replacing a complex part of the equation with a simpler variable, to make the equation easier to solve.
In our exercise, we start with the equation:
\( x^{6} - 10x^{3} = -9 \)
This equation is hard to solve directly because of the high degree of the polynomial. To simplify, we substitute
\( y = x^{3} \)
This reduces our equation to a simpler form:
\( y^{2} - 10y = -9 \)
Substitution helps break down the problem into more manageable steps. Once we solve the simpler equation, we can substitute back to find the original variable.
Quadratic Equations
Quadratic equations are polynomials of degree 2 and have the general form:
\[ ax^{2} + bx + c = 0 \]
In our example, after substitution, our equation turns into a quadratic equation:
\( y^{2} - 10y + 9 = 0 \)
Quadratic equations can be solved using various methods, including:
  • Factoring
  • Quadratic formula: \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \
  • Completing the square

    • Using any of these methods, we can find the values of the variable that make the equation true. In this exercise, we will factor the quadratic equation to find the roots.
Factoring
Factoring is breaking down a complex expression into simpler multipliers that, when multiplied together, give the original expression. For our quadratic equation:
\( y^{2} - 10y + 9 = 0 \)
The factors of this equation are:
\( (y - 1)(y - 9) = 0 \)
To find the roots:
\( y - 1 = 0 \) or \( y - 9 = 0 \)
Therefore,
\( y = 1 \) or \( y = 9 \)
Once factored, solving the equation is straightforward. After finding the values of y, we substitute back to find the values of x. For example,
if \( y = 1 \, \ x^{3} = 1 \ implies \ x = 1 \), and if \( y = 9 \, \ x^{3} = 9 \ implies \ x = \root{3}{9} \)
Factoring is essential for solving polynomial equations efficiently.

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