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Magazine Chapter 9: Problem 157
Solve. $$x^{4}-13 x^{2}-30=0$$
Short Answer
Expert verified
The real solutions are \( x = \sqrt{15} \) and \( x = -\sqrt{15} \).
Step by step solution
01
Substitute Variable
Let us substitute a new variable. Let us set \[ y = x^2 \] This turns the equation \( x^4 - 13x^2 - 30 = 0 \) into a quadratic equation: \[ y^2 - 13y - 30 = 0 \]
02
Solve the Quadratic Equation
To solve the quadratic equation \( y^2 - 13y - 30 = 0 \), we can use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -13 \), and \( c = -30 \). Plugging in these values:\[ y = \frac{13 \pm \sqrt{13^2 - 4(1)(-30)}}{2(1)} \] \[ y = \frac{13 \pm \sqrt{169 + 120}}{2} \] \[ y = \frac{13 \pm \sqrt{289}}{2} \] \[ y = \frac{13 \pm 17}{2} \]
03
Simplify the Roots
Simplifying the roots, we get two values for \( y \): \[ y = \frac{30}{2} = 15 \] \[ y = \frac{-4}{2} = -2 \]
04
Substitute Back
Recall that \( y = x^2 \). So, now substitute back to solve for \( x \). We have two equations: \[ x^2 = 15 \] \[ x^2 = -2 \]
05
Solve for x
For \( x^2 = 15 \), we get: \[ x = \pm \sqrt{15} \] For \( x^2 = -2 \), there are no real solutions because the square root of a negative number is imaginary.
06
Final Answer
Hence, the real solutions for the equation \( x^4 - 13x^2 - 30 = 0 \) are: \[ x = \sqrt{15} \] \[ x = -\sqrt{15} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Substitution
Variable substitution is a powerful tool in algebra, especially when dealing with complex equations. By replacing a variable or expression with another, simpler form, we can make an equation easier to solve.
In the given problem, we start with the equation \[ x^4 - 13x^2 - 30 = 0 \]
To simplify, we use the substitution \[ y = x^2 \]
This changes our equation from a quartic equation to a quadratic one:
In the given problem, we start with the equation \[ x^4 - 13x^2 - 30 = 0 \]
To simplify, we use the substitution \[ y = x^2 \]
This changes our equation from a quartic equation to a quadratic one:
- Original: \[ x^4 - 13x^2 - 30 = 0 \]
- New: \[ y^2 - 13y - 30 = 0 \]
Quadratic Formula
The quadratic formula \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] is a reliable method to find the roots of any quadratic equation \[ y^2 + by + c = 0 \].
For our problem, substituting \[ a = 1 \], \[ b = -13 \], and \[ c = -30 \]:
\[ y = \frac{13 \pm \sqrt{13^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \]
\[ y = \frac{13 \pm \sqrt{169 + 120}}{2} \]
\[ y = \frac{13 \pm \sqrt{289}}{2} \]
\[ y = \frac{13 \pm 17}{2} \]
This results in two potential solutions: \[ y = 15 \] and \[ y = -2 \].
For our problem, substituting \[ a = 1 \], \[ b = -13 \], and \[ c = -30 \]:
\[ y = \frac{13 \pm \sqrt{13^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \]
\[ y = \frac{13 \pm \sqrt{169 + 120}}{2} \]
\[ y = \frac{13 \pm \sqrt{289}}{2} \]
\[ y = \frac{13 \pm 17}{2} \]
This results in two potential solutions: \[ y = 15 \] and \[ y = -2 \].
Real and Imaginary Solutions
When solving quadratic equations, you may encounter both real and imaginary solutions.
By substituting back \[ y = x^2 \] into our solutions, we find:
By substituting back \[ y = x^2 \] into our solutions, we find:
- For \[ y = 15 \], \[ x^2 = 15 \]. This results in \[ x = \pm \sqrt{15} \], which are real solutions.
- For \[ y = -2 \], \[ x^2 = -2 \]. As taking square roots of negative numbers involve imaginary numbers, we say this has no real solutions.
Intermediate Algebra Problem Solving
Solving problems in intermediate algebra often requires a mix of several strategies. Here's a brief recap of the elements we used:
- **Variable Substitution**: Simplifies more complex equations into more manageable forms, like turning a quartic into a quadratic.
- **Quadratic Formula**: Offers a way to systematically solve quadratic equations, ensuring all possible roots are found.
- **Discriminating between Real and Imaginary Solutions**: Helps determine which solutions fit the context of the problem.
