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Quadratic substitution is a helpful technique for solving polynomial equations that aren't in simple quadratic form. It can make complex equations easier to manage. In this exercise, we used quadratic substitution by letting \( y = x^2 \). This substitution transforms the original polynomial equation, \( x^4 - 5x^2 + 4 = 0 \), into a simpler quadratic equation, \( y^2 - 5y + 4 = 0 \). This makes the equation easier to solve, as we can now use methods suited for quadratic equations. Without this substitution, directly solving the quartic equation might be more cumbersome. Remember, the key step here is recognizing when an equation can be simplified by substituting a suitable variable.

Factoring quadratic equations is a powerful method for finding solutions to these equations. Once we performed the substitution and transformed our equation into \( y^2 - 5y + 4 = 0 \), the next step was to factor it. Factoring is the process of breaking down the quadratic equation into simpler expressions that can be multiplied to give the original quadratic. For our equation, this meant finding two numbers that multiply to 4 and add up to -5. These numbers are -1 and -4, so we can write \( y^2 - 5y + 4 \) as \( (y - 1)(y - 4) \). This step allows us to set each factor equal to zero and solve for y, giving us \( y = 1 \) and \( y = 4 \). Factoring quadratics often serves as a critical step in solving these types of equations.

After factoring the quadratic equation and solving for y, the next task is to find the real number solutions for the original variable. Recall that we substituted \( y = x^2 \). So, we substitute back to get \( x^2 = 1 \) and \( x^2 = 4 \). To find the real solutions for x, we solve these equations: \( x^2 = 1 \) gives us \( x = \pm 1 \), and \( x^2 = 4 \) gives us \( x = \pm 2 \). These are the final real solutions to the original equation \( x^4 - 5x^2 + 4 = 0 \). Remember, real number solutions are simply the values of x that satisfy the given polynomial equation and are real numbers (not involving imaginary or complex numbers). Identifying these solutions involves back-substituting and solving the resulting simpler equations.