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The process of transforming a quartic equation into a quadratic equation is called quadratic transformation. In our problem, the equation starts as a quartic, which is an equation of the fourth degree: \(x^4 + x - 3 = 0\). When tackling quartic equations, sometimes they can be simplified by eliminating certain powers of \(x\) with clever substitutions.

This transformation relies on recognizing patterns within the equation. Here, the idea is to simplify \(x^4\) using the substitution \(u = x^2\). This effectively reduces the problem by half to a quadratic form. This kind of transformation leverages our deeper understanding of algebra and equation forms to make a problem more manageable, paving the way for simpler methods to solve it.

Variable substitution is a technique used to simplify complicated equations. This method involves replacing an existing variable with a new one, making the equation easier to handle. In the given exercise, the substitution \(u = x^2\) cleverly simplifies \(x^4\) to \(u^2\).

By making this substitution, the entire problem becomes a quadratic equation: \(u^2 + u - 3 = 0\). This new form is familiar and more straightforward to solve. Variable substitution can also help in visualizing the problem differently, where each variable represents a more familiar part of the equation, thereby simplifying complex algebra into recognizable forms.

Once we have transformed the quartic equation into a quadratic equation via substitution, the next step is solving it. Quadratic equations like \(u^2 + u - 3 = 0\) can be solved using the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

For this equation, \(a = 1\), \(b = 1\), and \(c = -3\). This gives us the solutions for \(u\). After calculating using the formula, we find two solutions: \(u = -2\) and \(u = 1.5\). Remember, it's crucial to learn the quadratic formula as it's a powerful tool that provides an exact solution to any quadratic equation, assisting in breaking down more complex problems like quartic equations.

Finding the roots of an equation means deducing the values of the variable that satisfy the equation. In our transformed quadratic equation, we found \(u\) values which served as solutions. However, since the substitution \(u = x^2\) was employed, we must translate these solutions back to \(x\).

For \(u = -2\), there are no real solutions for \(x\) since no real number squared gives a negative number. However, for \(u = 1.5\), \(x \) can be derived as \(x = \pm \sqrt{1.5}\). Thus, the real roots of the original quartic equation are \(+\sqrt{1.5}\) and \(-\sqrt{1.5}\). These roots are the values of \(x\) that solve the original equation, learned through transforming and solving a simpler quadratic form.