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When tackling a quartic polynomial, like the one in our exercise: \( x^4 - 5x^2 + 4 \), solving directly can be a bit complex. Instead, we can leverage the idea of quadratic substitution, which simplifies our task significantly. The equation \( x^4 - 5x^2 + 4 \) can be transformed using a clever substitution. By letting \( y = x^2 \), the quartic equation converts into a much simpler quadratic form: \( y^2 - 5y + 4 = 0 \). This substitution is incredibly useful because it temporarily reduces the degree of the polynomial we're working with. Instead of dealing with powers of four, we're now looking at a manageable quadratic problem.

• Substitute \( y = x^2 \) to change a quartic to a quadratic.
• Simplifies the problem, making root-finding straightforward.

This method is particularly powerful because it transforms the task into something we may have already mastered: solving quadratics. The simplicity introduced by substitution allows us to proceed with confidence onto the next steps.

Once we've transformed our quartic expression into a quadratic equation, we can solve it using the quadratic formula. This formula is a reliable tool derived from the process of completing the square and provides the solutions to any quadratic equation of the form \( ax^2 + bx + c = 0 \).For our equation \( y^2 - 5y + 4 = 0 \), we set:- \( a = 1 \)- \( b = -5 \)- \( c = 4 \)Then, applying the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute these values in:- Calculate \( b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 4 = 25 - 16 = 9 \).- Find \( y = \frac{5 \pm \sqrt{9}}{2} \).

• Simplifies to \( y = \frac{5 \pm 3}{2} \).
• Results are \( y = 4 \) and \( y = 1 \).

The formula provides both roots of the quadratic equation, which pave the way for solving our initial quartic problem.

The step of root finding begins once we have the solutions from the quadratic equation. In our scenario, substituting back from \( y \) to \( x \) is crucial to unravel the original equation's roots.Our solutions from \( y^2 - 5y + 4 = 0 \) were \( y = 4 \) and \( y = 1 \). Now, recall:- If \( y = x^2 \), then for \( y = 4 \), \( x^2 = 4 \).- Solving for \( x \) yields \( x = \pm 2 \).- Similarly, for \( y = 1 \), \( x^2 = 1 \).- Thus, \( x = \pm 1 \).Therefore, the roots of the original quartic equation \( x^4 - 5x^2 + 4 = 0 \) are \( x = 2, -2, 1, -1 \).

• Utilize found \( y \) values to find all possible \( x \) roots.
• Recognize that both positive and negative solutions exist for each \( x^2 \).

Understanding and executing this root finding maximizes our problem-solving strategy by ensuring all potential solutions are considered.