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Magazine Chapter 11: Problem 74
Solve. $$ x^{6}+7 x^{3}-8=0 $$
Short Answer
Expert verified
The solutions are \( x = -2 \) and \( x = 1 \).
Step by step solution
01
Substitute Variable
Let鈥檚 substitute a new variable to make the equation simpler. Set \( y = x^3 \). Hence, the equation \( x^6 + 7x^3 - 8 = 0 \) can be rewritten as \( y^2 + 7y - 8 = 0 \).
02
Solve the Quadratic Equation
Solve the quadratic equation \( y^2 + 7y - 8 = 0 \) by factoring. We need two numbers that multiply to -8 and add up to 7. The numbers are 8 and -1. Therefore, \( y^2 + 7y - 8 = (y + 8)(y - 1) = 0 \).
03
Find the Roots of the Quadratic Equation
Set each factor equal to zero: \( y + 8 = 0 \) and \( y - 1 = 0 \). Thus, the solutions are \( y = -8 \) and \( y = 1 \).
04
Substitute Back the Original Variable
Substitute back \( y = x^3 \) into the solutions. So, we get \( x^3 = -8 \) and \( x^3 = 1 \).
05
Solve for x
Now solve for \( x \). For \( x^3 = -8 \), we have \( x = \sqrt[3]{-8} = -2 \). For \( x^3 = 1 \), we have \( x = \sqrt[3]{1} = 1 \).
06
Final Answer
The solutions to the equation \( x^6 + 7x^3 - 8 = 0 \) are \( x = -2 \) and \( x = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
substitute variable
One key strategy in solving complex polynomial equations is to 'substitute a variable.' Essentially, we replace a part of the equation with a simpler variable to make it easier to handle. In this case, we substitute 聽\( y = x^3 \). This helps transform our given equation \( x^6 + 7x^3 - 8 = 0 \) into a simpler quadratic form: \( y^2 + 7y - 8 = 0 \).
This substitution makes the equation manageable and straightforward to solve. For instance, instead of dealing with powers higher than two, we now work with a basic quadratic equation.
This substitution makes the equation manageable and straightforward to solve. For instance, instead of dealing with powers higher than two, we now work with a basic quadratic equation.
quadratic equation
A quadratic equation is a fundamental concept of algebra in the form 聽\( ax^2 + bx + c = 0 \). The equation we derived from our substitution is \( y^2 + 7y - 8 = 0 \). Here, we identify that:
Quadratic equations often have two solutions, corresponding to the positive and negative roots of the quadratic formula.
- 聽\( a = 1 \)
- \( b = 7 \)
- \( c = -8 \)
Quadratic equations often have two solutions, corresponding to the positive and negative roots of the quadratic formula.
factoring
Factoring is a method used to solve quadratic equations by expressing them as the product of two binomials. For the equation 聽\( y^2 + 7y - 8 = 0 \), we look for two numbers that multiply to -8 and add up to 7.
These numbers are 8 and -1, hence we can factorize as follows:
These numbers are 8 and -1, hence we can factorize as follows:
- \( y^2 + 7y - 8 = (y + 8)(y - 1) \)
finding roots
Finding the roots means finding the values that make the equation zero. From our factored form \((y + 8)(y - 1) = 0 \), we can set each factor to zero to find the roots:first step:
To find the actual solutions in terms of x, we need to substitute back. This involves reversing our substitution step to 聽\( y = x^3 \), leading to the next concept.
- \( y + 8 = 0 \ y = -8 \)
- \( y - 1 = 0 \ y = 1 \)
To find the actual solutions in terms of x, we need to substitute back. This involves reversing our substitution step to 聽\( y = x^3 \), leading to the next concept.
cubing
Cubing, or solving for the cubic root, is the final step in our solution. We have the roots for 聽\( y \)聽 as -8 and 1. Substituting 聽\( x^3 = y \) back, we get:
Understanding cubing in this context helps to complete the loop, bringing us back to solving the polynomial equation from where we started.
- \( x^3 = -8 \ x = \,^{3}{^-8} = -2 \)
- \( x^3 = 1 \ x = \,^{3}{^1}=1 \)
Understanding cubing in this context helps to complete the loop, bringing us back to solving the polynomial equation from where we started.
