Problem 41 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ 2 x^{4}+4 x^{2}+1=0 $$

Step-by-Step Solution

Answer

There are no real solutions for the equation given.

Step 1: Understand the Form of the Equation

The given equation is a polynomial equation of degree 4: \(2x^4 + 4x^2 + 1 = 0\). Notice that it is a quadratic in form if we let \(y = x^2\), thus treating the equation as \(2y^2 + 4y + 1 = 0\).

Step 2: Change of Variable

Let \(y = x^2\). Then the equation becomes \(2y^2 + 4y + 1 = 0\). This is now a quadratic equation in \(y\).

Step 3: Apply the Quadratic Formula

The quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) will be used here with \(a = 2\), \(b = 4\), and \(c = 1\).

Step 4: Compute the Discriminant

Calculate the discriminant: \(b^2 - 4ac = 4^2 - 4(2)(1) = 16 - 8 = 8\). Since the discriminant is positive, there are two distinct real solutions for \(y\).

Step 5: Solve for y Using Quadratic Formula

Substitute into the quadratic formula: \(y = \frac{-4 \pm \sqrt{8}}{4}\). Simplify \(\sqrt{8}\) to \(2\sqrt{2}\), resulting in \(y = \frac{-4 \pm 2\sqrt{2}}{4}\). Simplify to get \(y = -1 \pm \frac{\sqrt{2}}{2}\).

Step 6: Reinterpret Solutions in Terms of x

Recall \(y = x^2\). Therefore, set up the equations \(x^2 = -1 + \frac{\sqrt{2}}{2}\) and \(x^2 = -1 - \frac{\sqrt{2}}{2}\).

Step 7: Solve for x

Check the values of \(x^2\): \(-1 + \frac{\sqrt{2}}{2}\) and \(-1 - \frac{\sqrt{2}}{2}\). Both expressions are negative, indicating no real solutions for \(x^2\). Thus, there are no real solutions for \(x\).

Key Concepts

Quartic Equations

Quartic equations are polynomial equations of degree four. These equations can look complex at first glance, but they often hold interesting patterns that can simplify the solving process. They take the general form \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \]where the highest exponent is four. In our original exercise, the equation is \[ 2x^4 + 4x^2 + 1 = 0. \]Because the term with the third power, \(x^3\), is missing, it hints that there might be a simpler way to handle this problem. Applying a few algebraic tricks like factoring or substitution can turn this complex quartic into something more manageable. This specific quartic equation is structured in a way that allows us to express it in terms of a simpler quadratic form.

Quadratic in Form

A quartic equation can occasionally be quadratic in form, meaning it can be rewritten such that it resembles a standard quadratic equation. Recognizing this structure is critical for simplifying and solving the equation. In the problem \(2x^4 + 4x^2 + 1 = 0\), notice the similarity between the powers of \(x\). You can think of \(x^4\) as \((x^2)^2\), allowing you to perform a substitution to make the equation look like a quadratic: \[ 2(x^2)^2 + 4x^2 + 1 = 0. \]By setting \(y = x^2\), the equation becomes \[ 2y^2 + 4y + 1 = 0, \] which is clearly a quadratic in \(y\). This transformation makes it considerably simpler to use familiar quadratic methods to find the solutions to the equation.

Quadratic Formula

Once we have a quadratic in the form \(ay^2 + by + c = 0\), the quadratic formula becomes a powerful tool for finding solutions. The formula is given by:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]which provides two potential solutions depending on whether you add or subtract the square root term.For the quadratic \(2y^2 + 4y + 1 = 0\), we identify:

• \(a = 2\)
• \(b = 4\)
• \(c = 1\)

We can then plug these values into the formula. Begin by calculating the discriminant, \(b^2 - 4ac\). A positive discriminant implies the presence of two real and distinct solutions for \(y\). In our case, \[ b^2 - 4ac = 16 - 8 = 8, \]yielding a positive number, thus confirming two distinct real solutions for \(y\).

Real Solutions

After applying the quadratic formula, we determined the solutions for \(y\) are \[ y = -1 \pm \frac{\sqrt{2}}{2}. \]These solutions lead us back to \(x\) because we previously set \(y = x^2\). So, our next step would be to set \[ x^2 = -1 + \frac{\sqrt{2}}{2} \]and \[ x^2 = -1 - \frac{\sqrt{2}}{2}. \]For a solution to be real, \(x^2\) must be a non-negative number. However, both expressions for \(x^2\) in this case turn out to be negative. Therefore,

• There are no real values of \(x\) that satisfy these equations,
• Implying there are no real solutions for the original quartic equation.

This shows the importance of verifying whether the solutions are real after finding them through manipulation and transformation.