Problem 82
Question
The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(8 x^{4}+1=11 x^{2}\)
Step-by-Step Solution
Verified Answer
The solutions are \( x = \pm \sqrt{\frac{11 + \sqrt{89}}{16}} \) and \( x = \pm \sqrt{\frac{11 - \sqrt{89}}{16}} \).
1Step 1: Bring all terms to one side of the equation
Start by moving all terms to one side of the equation so it equals zero: \[ 8x^4 + 1 - 11x^2 = 0 \]
2Step 2: Use substitution
To simplify, use the substitution \( y = x^2 \). Then the equation becomes: \[ 8y^2 - 11y + 1 = 0 \]
3Step 3: Solve the quadratic equation
Recognize this as a quadratic equation in terms of \( y \). Use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 8 \), \( b = -11 \), and \( c = 1 \). Substitute these values:\[ y = \frac{11 \pm \sqrt{121 - 32}}{16} = \frac{11 \pm \sqrt{89}}{16} \]
4Step 4: Back-substitute \( y = x^2 \)
Since we know that \( y = x^2 \), set \( x^2 \) to each of the values found for \( y \):\[ x^2 = \frac{11 + \sqrt{89}}{16} \] and \[ x^2 = \frac{11 - \sqrt{89}}{16} \]
5Step 5: Solve for \( x \)
Take the square root of both sides for each case:Case 1: \[ x = \pm \sqrt{\frac{11 + \sqrt{89}}{16}} \]Case 2: \[ x = \pm \sqrt{\frac{11 - \sqrt{89}}{16}} \]
Key Concepts
Quadratic SubstitutionQuadratic FormulaComplex Numbers
Quadratic Substitution
Solving a quartic equation like the one given in the exercise can sometimes be simplified using a method called quadratic substitution.
This involves substituting a new variable to make the equation easier to handle. In this problem, we notice that there's a mix of powers of 4 and powers of 2.
This suggests that setting a substitution where the quadratic terms become simpler terms will be helpful.
We start with the equation:\[ 8x^4 + 1 - 11x^2 = 0 \]
By setting \( y = x^2 \), we reduce the complexity of the problem. The equation then becomes:\[ 8y^2 - 11y + 1 = 0 \]
Now, what was once a complicated quartic equation is now a straightforward quadratic equation.
Quadratic substitution doesn't change the solution—it's just a clever way to make the math more manageable. It’s a useful trick when dealing with higher-degree polynomials.
This substitution step paves the way for using the quadratic formula.
This involves substituting a new variable to make the equation easier to handle. In this problem, we notice that there's a mix of powers of 4 and powers of 2.
This suggests that setting a substitution where the quadratic terms become simpler terms will be helpful.
We start with the equation:\[ 8x^4 + 1 - 11x^2 = 0 \]
By setting \( y = x^2 \), we reduce the complexity of the problem. The equation then becomes:\[ 8y^2 - 11y + 1 = 0 \]
Now, what was once a complicated quartic equation is now a straightforward quadratic equation.
Quadratic substitution doesn't change the solution—it's just a clever way to make the math more manageable. It’s a useful trick when dealing with higher-degree polynomials.
This substitution step paves the way for using the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ay^2 + by + c = 0 \). It's given by:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are constants from the quadratic equation. In our simplified equation:\[ 8y^2 - 11y + 1 = 0 \]
we can identify \( a = 8 \), \( b = -11 \), and \( c = 1 \).
Plugging these values into the quadratic formula, we get:\[ y = \frac{11 \pm \sqrt{121 - 32}}{16} = \frac{11 \pm \sqrt{89}}{16} \](since \( b = -11 \), the formula simplifies to \( 11 \) instead of \( -(-11) \)).
This formula step provides two potential solutions for \( y \). These solutions are critical in the next phase where we revert our substitution.
we can identify \( a = 8 \), \( b = -11 \), and \( c = 1 \).
Plugging these values into the quadratic formula, we get:\[ y = \frac{11 \pm \sqrt{121 - 32}}{16} = \frac{11 \pm \sqrt{89}}{16} \](since \( b = -11 \), the formula simplifies to \( 11 \) instead of \( -(-11) \)).
This formula step provides two potential solutions for \( y \). These solutions are critical in the next phase where we revert our substitution.
Complex Numbers
As we process the solutions of our quadratic equation, we end up with two values for \( y \):\[ y = \frac{11 + \sqrt{89}}{16} \] and\[ y = \frac{11 - \sqrt{89}}{16} \]Step 4 tells us to back-substitute \( y = x^2 \). Hence, we equate \( x^2 \) to these \( y \) values and proceed to solve for \( x \).
While often you may end with real numbers, sometimes the solutions may include complex numbers.
The term \( \sqrt{89} \) is irrational but real because 89 is a positive number. Therefore, the roots derived from our quadratic will be real numbers. However, if we had an equation where the discriminant \( b^2 - 4ac \) was negative, our solutions would involve complex numbers.
In such cases, we'd use the imaginary unit \( i \), defined as \( \sqrt{-1} \), to express our solutions. For instance, if our discriminant were negative, say \( -89 \), we would have:\[ y = \frac{11 \pm \sqrt{-89}}{16} \] and thus:\[ y = \frac{11 \pm i\sqrt{89}}{16}\]Although this example uses real numbers, understanding complex numbers is crucial for a comprehensive grasp of all possible quadratic solutions.
While often you may end with real numbers, sometimes the solutions may include complex numbers.
The term \( \sqrt{89} \) is irrational but real because 89 is a positive number. Therefore, the roots derived from our quadratic will be real numbers. However, if we had an equation where the discriminant \( b^2 - 4ac \) was negative, our solutions would involve complex numbers.
In such cases, we'd use the imaginary unit \( i \), defined as \( \sqrt{-1} \), to express our solutions. For instance, if our discriminant were negative, say \( -89 \), we would have:\[ y = \frac{11 \pm \sqrt{-89}}{16} \] and thus:\[ y = \frac{11 \pm i\sqrt{89}}{16}\]Although this example uses real numbers, understanding complex numbers is crucial for a comprehensive grasp of all possible quadratic solutions.
Other exercises in this chapter
Problem 81
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