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Magazine Chapter 4: Problem 19
Find the four real zeros of the function \(f(x)=2 x^{4}-4 x^{2}+1\)
Short Answer
Expert verified
The zeros are \(\pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(\pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).
Step by step solution
01
Identify the Type of Equation
The given function is a polynomial of degree 4: \(f(x) = 2x^4 - 4x^2 + 1\). It is a quartic equation, which can sometimes be solved by substitution, especially if it looks like a quadratic in form.
02
Substitute to Simplify the Equation
To simplify the equation, notice that it can be treated like a quadratic in terms of \(x^2\). Let \(u = x^2\), then the equation becomes \(2u^2 - 4u + 1 = 0\).
03
Solve the Quadratic Equation
Now solve the quadratic equation \(2u^2 - 4u + 1 = 0\). Use the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -4\), and \(c = 1\).
04
Calculate the Discriminant
Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 2 \times 1 = 16 - 8 = 8\). Since the discriminant is positive, there are two real solutions for \(u\).
05
Find the Values of \(u\)
Substitute the discriminant back into the quadratic formula: \(u = \frac{4 \pm \sqrt{8}}{4}\). Simplify \(\sqrt{8} = 2\sqrt{2}\), so: \(u = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}\). The solutions for \(u\) are \(1 + \frac{\sqrt{2}}{2}\) and \(1 - \frac{\sqrt{2}}{2}\).
06
Determine the Values of \(x\)
Convert back to \(x\) with \(x^2 = u\). Therefore, solve \(x^2 = 1 + \frac{\sqrt{2}}{2}\) and \(x^2 = 1 - \frac{\sqrt{2}}{2}\). Take the square root of both sides to find \(x\).
07
Compute the Four Zeros
For \(x^2 = 1 + \frac{\sqrt{2}}{2}\), \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\). For \(x^2 = 1 - \frac{\sqrt{2}}{2}\), \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\). Therefore, the four zeros are: \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations form a fundamental part of algebra and are typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The equation in this problem, after the substitution \(u = x^2\), transforms a quartic equation into a quadratic one. This process highlights that even higher-degree polynomials can sometimes be reduced to quadratic equations.
Quadratic equations are solvable by various methods, such as:
Quadratic equations are solvable by various methods, such as:
- Factoring, when the equation can be expressed as a product of linear factors.
- Completing the square, a technique that involves adding a term to both sides to form a perfect square trinomial.
- Quadratic formula, given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), useful when the equation does not easily factorize.
Quartic Equations
A quartic equation is a polynomial equation of degree four, represented as \(ax^4 + bx^3 + cx^2 + dx + e = 0\). These equations can have up to four real or complex roots and may initially seem intimidating.
In the given exercise, the equation \(f(x) = 2x^4 - 4x^2 + 1\) is a quartic equation. This specific type can often be simplified by substitution, turning it into a quadratic equation in terms of another variable.
Some common strategies to solve quartic equations include:
In the given exercise, the equation \(f(x) = 2x^4 - 4x^2 + 1\) is a quartic equation. This specific type can often be simplified by substitution, turning it into a quadratic equation in terms of another variable.
Some common strategies to solve quartic equations include:
- Substitution, which reduces the complexity by introducing a new variable.
- Factoring, if the polynomial is factorable into lower-degree polynomials.
- Using methods developed specifically for quartic equations, such as Ferrari鈥檚 method.
Discriminant
The discriminant is a significant feature of quadratic equations, providing insight into the nature and number of roots without solving the equation. It is calculated as \(b^2 - 4ac\). The discriminant helps to identify if roots are real or complex, and also if they are distinct or repeated:
- If the discriminant is positive, the equation has two distinct real roots.
- If it is zero, there is exactly one real root, repeated.
- If negative, the roots are complex (not real).
Real Zeros
Identifying the real zeros of a polynomial function involves finding the values of \(x\) where the function \(f(x)\) equals zero. These values are significant as they represent the x-intercepts of the polynomial graph.
The zeros in this instance were obtained by solving \(x^2 = u\), where \(u\) was previously found using the quadratic formula. For the function \(f(x) = 2x^4 - 4x^2 + 1\), the zeros are \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).
Steps to find Real Zeros:
The zeros in this instance were obtained by solving \(x^2 = u\), where \(u\) was previously found using the quadratic formula. For the function \(f(x) = 2x^4 - 4x^2 + 1\), the zeros are \(x = \pm \sqrt{1 + \frac{\sqrt{2}}{2}}\) and \(x = \pm \sqrt{1 - \frac{\sqrt{2}}{2}}\).
Steps to find Real Zeros:
- Transform the quartic equation to quadratic form by an appropriate substitution.
- Solve for variable using the quadratic formula or other suitable method.
- Revert to the original variable to determine real solutions or zeros.
