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Magazine Chapter 14: Problem 7
Solve the given equations algebraically. In Exercise \(10,\) explain your method. $$x^{-4}+2 x^{-2}=24$$
Short Answer
Expert verified
The solutions are \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).
Step by step solution
01
Substitution
To simplify the given equation, we'll use substitution. Let \( y = x^{-2} \). This implies that \( x^{-4} = y^2 \). Rewriting the equation with these substitutions gives us: \[ y^2 + 2y = 24 \]
02
Rearrange into a Quadratic Equation
To solve for \( y \), we need to rearrange the equation into a standard quadratic form. Thus, subtract 24 from both sides: \[ y^2 + 2y - 24 = 0 \] This is now a quadratic equation in terms of \( y \).
03
Solve the Quadratic Equation
We solve the quadratic equation \( y^2 + 2y - 24 = 0 \). It can be factored as \[ (y + 6)(y - 4) = 0 \] Therefore, \( y + 6 = 0 \) or \( y - 4 = 0 \). Solving these gives \( y = -6 \) or \( y = 4 \).
04
Substitute Back to Find x
Recall that \( y = x^{-2} \). Therefore, if \( y = -6 \), then \( x^{-2} = -6 \). Since a negative value cannot equal \( x^{-2} \), we discard this solution. If \( y = 4 \), then we solve \[ x^{-2} = 4 \] which implies \[ x^2 = rac{1}{4} \] Solving for \( x \), we get \( x = rac{1}{2} \) or \( x = - rac{1}{2} \).
05
Verify the Solutions
Verify by plugging \( x = rac{1}{2} \) and \( x = - rac{1}{2} \) back into the original equation:- For \( x = rac{1}{2} \): \( x^{-4} = 16 \) and \( 2x^{-2} = 8 \). Thus, total is \( 24 \).- For \( x = - rac{1}{2} \): \( x^{-4} = 16 \) and \( 2x^{-2} = 8 \). Thus, total is \( 24 \).Both values satisfy the equation. Thus, both are solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are algebraic expressions of the form \( ax^2 + bx + c = 0 \). They are called 'quadratic' because the highest power of the variable is 2. In solving quadratic equations, our main goal is to find the values of \( x \) that make the equation true. These values are called the roots of the equation.
- Standard form: This is how we express quadratic equations, where \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \).
- Solutions: We often find two solutions for \( x \) because the variable \( x \) is squared.
Substitution Method
The substitution method is a powerful algebraic tool that simplifies complex equations by replacing a variable with another expression. In the original exercise, we used substitution to transform a complex equation with negative exponents into a simpler quadratic form.
- Step 1: Identify substitution: In the equation \( x^{-4} + 2x^{-2} = 24 \), let \( y = x^{-2} \). This reduces the complexity of the equation.
- Step 2: Replace in the equation: The substitution changes the form to \( y^2 + 2y = 24 \).
Solving Algebraic Equations
Solving algebraic equations is about finding the values of the variables that satisfy the equation. Here, the process involved multiple strategies to arrive at the final answer.
- Step 1: Rearranging: After substituting, we aim to bring the equation to a standard form. The equation \( y^2 + 2y = 24 \) is rewritten as \( y^2 + 2y - 24 = 0 \).
- Step 2: Factoring: This equation is factored as \( (y + 6)(y - 4) = 0 \). This gives two possible values for \( y \).
- Step 3: Solving for x: Substituting back, we find the corresponding \( x \) values by solving \( x^{-2} = y \). In this case, \( x^2 = \frac{1}{4} \) suggests \( x = \frac{1}{2} \) or \( x = -\frac{1}{2} \).
Verification of Solutions
Verification of solutions is a crucial step in solving algebraic equations. It ensures that the solutions computed are valid and satisfy the original equation.
- Check substitution: Return to the original equation with the solution values. For \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \), both satisfy \( x^{-4} + 2x^{-2} = 24 \).
- Validate equalities: Substituting back into \( x^{-4} \) and \( 2x^{-2} \), verify each term and ensure their sum equals 24.
