91Ó°ÊÓ

Quadratic substitution is a technique used to simplify complex equations and make them easier to solve. In the given problem, we start with the equation: \[2x^4 + 4x^2 + 1 = 0\]At first glance, this equation may seem difficult to solve directly. However, by recognizing patterns, we can simplify it. Notice that the terms involve powers of 2, specifically \(x^4\) and \(x^2\).
To transform the equation into an easier form, we assume \(y = x^2\). Substituting \(y\) for \(x^2\) transforms the original equation into:\[2y^2 + 4y + 1 = 0\]This is a standard quadratic equation in terms of \(y\). By substituting a single variable for \(x^2\), the complexity of the equation is reduced, making it manageable with basic algebraic techniques.
In essence, quadratic substitution helps you see the problem from a different perspective, focusing on simpler components rather than complicated higher powers.

The quadratic formula is a universal tool for solving any quadratic equation. Given a quadratic equation in the form \(ax^2 + bx + c = 0\), it provides a straightforward way to find the roots. The formula is:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our quadratic equation \(2y^2 + 4y + 1 = 0\), we identify:

• \(a = 2\)
• \(b = 4\)
• \(c = 1\)

Plug these values into the quadratic formula to get:\[y = \frac{-4 \pm \sqrt{4^2 - 4(2)(1)}}{4}\]Simplifying further:\[y = \frac{-4 \pm \sqrt{16 - 8}}{4} = \frac{-4 \pm \sqrt{8}}{4}\]Breaking it down, and reducing \(\sqrt{8}\) to \(2\sqrt{2}\), the solutions are:\[y_1 = \frac{-2 + \sqrt{2}}{2},\ y_2 = \frac{-2 - \sqrt{2}}{2}\]Using the quadratic formula ensures accurate solutions, even for equations that are hard to factor by other means.

When dealing with quadratic equations in terms of \(x^2\), the solutions must be real and non-negative. Since \(x^2\) represents the square of a real number, \(y = x^2\) must be non-negative. Thus, we must verify whether the solutions derived are non-negative.
In our case, the solutions were:

• \(y_1 = \frac{-2 + \sqrt{2}}{2}\)
• \(y_2 = \frac{-2 - \sqrt{2}}{2}\)

These need checking for non-negativity:- Calculate \(y_1\) approximately: \(\sqrt{2} \approx 1.414\). - Therefore \(\frac{-2 + 1.414}{2} < 0\), indicating that \(y_1\) is negative.
Similarly, \(y_2\) is also negative:- \(\frac{-2 - 1.414}{2}\) clearly results in a negative value.Consequently, both derived \(y\) values fail the non-negativity test, implying no feasible real solutions for \(x\). In summary, ensuring all solutions are real and non-negative is essential when dealing with equations involving squares.