Problem 17
Question
Find the four real zeros of the function \(f(x)=2 x^{4}-4 x^{2}+1\).
Step-by-Step Solution
Verified Answer
The zeros are \( x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}} \) and \( x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}} \).
1Step 1: Recognize the Polynomial Type
The function \( f(x) = 2x^4 - 4x^2 + 1 \) is a quartic polynomial, specifically in the form of a quadratic in terms of \( x^2 \). This suggests we can use substitution to simplify the problem.
2Step 2: Substitute to Simplify
To simplify solving for zeros, let \( u = x^2 \). Therefore, the equation becomes \( 2u^2 - 4u + 1 = 0 \). Now, you have a quadratic equation in terms of \( u \).
3Step 3: Use the Quadratic Formula
For the quadratic equation \( 2u^2 - 4u + 1 = 0 \), apply the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 2 \), \( b = -4 \), and \( c = 1 \).
4Step 4: Calculate the Discriminant
First, find the discriminant \( b^2 - 4ac \): \((-4)^2 - 4 \times 2 \times 1 = 16 - 8 = 8\). The discriminant is positive, indicating two real and distinct solutions for \( u \).
5Step 5: Solve for \( u \)
Substitute the values into the quadratic formula: \[ u = \frac{-(-4) \pm \sqrt{8}}{2 \times 2} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2} \]. So, the solutions for \( u \) are \( u = 1 + \frac{\sqrt{2}}{2} \) and \( u = 1 - \frac{\sqrt{2}}{2} \).
6Step 6: Resubstitute for \( x \)
Replace \( u = x^2 \) back: Thus, \( x^2 = 1 + \frac{\sqrt{2}}{2} \) and \( x^2 = 1 - \frac{\sqrt{2}}{2} \).
7Step 7: Solve for \( x \)
For each \( x^2 \) value, solve for \( x \):1. \( x^2 = 1 + \frac{\sqrt{2}}{2} \Rightarrow x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}} \).2. \( x^2 = 1 - \frac{\sqrt{2}}{2} \Rightarrow x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}} \).
Key Concepts
Polynomial FactorizationQuadratic SubstitutionQuadratic FormulaDiscriminant Calculation
Polynomial Factorization
When solving quartic polynomials like \(f(x)=2x^4 - 4x^2 + 1\), factorization is a powerful tool. Factorization involves expressing a polynomial as a product of its factors, making it simpler to solve.
\( f(x) \) resembles a quadratic but in terms of \( x^2 \). This characteristic allows us to utilize substitution (as in quadratic form) to break down the polynomial.
When you reframe \(f(x)\) into a quadratic form, you facilitate easier zero-finding for functions that seem complex at first.
This approach can turn challenging polynomials into simpler equations, speeding up the factorization process.
\( f(x) \) resembles a quadratic but in terms of \( x^2 \). This characteristic allows us to utilize substitution (as in quadratic form) to break down the polynomial.
When you reframe \(f(x)\) into a quadratic form, you facilitate easier zero-finding for functions that seem complex at first.
This approach can turn challenging polynomials into simpler equations, speeding up the factorization process.
Quadratic Substitution
Quadratic substitution is a strategic technique to simplify and solve polynomials.
By substituting \( u = x^2 \), we reframe the quartic polynomial \(2x^4 - 4x^2 + 1\) into a straightforward quadratic \(2u^2 - 4u + 1\).
This transformation allows us to leverage familiar techniques to solve complex polynomial equations.
By substituting \( u = x^2 \), we reframe the quartic polynomial \(2x^4 - 4x^2 + 1\) into a straightforward quadratic \(2u^2 - 4u + 1\).
This transformation allows us to leverage familiar techniques to solve complex polynomial equations.
- Identify the polynomial structure
- Choose an appropriate substitution (e.g., \( u = x^2 \))
- Convert to a simple quadratic form
Quadratic Formula
The quadratic formula is a reliable method for solving quadratic equations of the form \(ax^2 + bx + c = 0\).
It enables finding zeros directly using: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For the substitution \(2u^2 - 4u + 1 = 0\), applying the formula helps determine the solutions for \(u\).
This universal formula is particularly useful when factorization is not obvious or easy.
Knowing the coefficients \(a, b, c\) of your quadratic lets you directly apply the formula and find solutions quickly.
It enables finding zeros directly using: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For the substitution \(2u^2 - 4u + 1 = 0\), applying the formula helps determine the solutions for \(u\).
This universal formula is particularly useful when factorization is not obvious or easy.
Knowing the coefficients \(a, b, c\) of your quadratic lets you directly apply the formula and find solutions quickly.
Discriminant Calculation
The discriminant \(b^2 - 4ac\) is a key part of the quadratic formula.
It reveals the nature of the roots of a quadratic equation.
For our equation, the calculation is: \((-4)^2 - 4 \times 2 \times 1 = 16 - 8 = 8\)
The discriminant reveals several properties:
It reveals the nature of the roots of a quadratic equation.
For our equation, the calculation is: \((-4)^2 - 4 \times 2 \times 1 = 16 - 8 = 8\)
The discriminant reveals several properties:
- Positive discriminant: two distinct real roots
- Zero discriminant: one real root (repeated)
- Negative discriminant: complex roots
Other exercises in this chapter
Problem 16
Sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with T
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Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $
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Show that if \(f^{\prime \prime}>0\) throughout an interval \([a, b],\) then \(f^{\prime}\) has at most one zero in \([a, b] .\) What if \(f^{\prime \prime}
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Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y-x^{4}-2 x^{2}-x^{2}\left(x^{2}-2\right)$$
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