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A quadratic equation is a type of polynomial equation. Specifically, it is a second-degree polynomial, which means the highest power of the variable is 2. In general form, it can be written as \[ ax^2 + bx + c = 0 \] where:

• \( a \), \( b \), and \( c \) are constants with \( a eq 0 \).
• \( x \) is the variable or unknown we are solving for.

Quadratic equations can look straightforward, but sometimes they appear in disguised forms. For example, in the exercise, although the equation is \( x^4 - 12x^2 + 11 = 0 \), it can be seen as a quadratic in disguise. We achieve this by recognizing that \( x^4 \) can be rewritten as \((x^2)^2\). Therefore, letting \( y = x^2 \), the equation simplifies to the quadratic form \( y^2 - 12y + 11 = 0 \). This technique allows us to apply methods for solving quadratic equations to higher-degree polynomial equations.

The quadratic formula is a powerful algebraic tool we use to find solutions of a quadratic equation. For any quadratic equation \( ax^2 + bx + c = 0 \), the roots of the equation can be found using:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where:

• \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation.
• The symbol \( \pm \) indicates there will generally be two solutions: one using addition and another using subtraction.

In our exercise, after substituting \( y = x^2 \), the simplified quadratic equation \( y^2 - 12y + 11 = 0 \) can be solved directly using the quadratic formula. By identifying \( a = 1 \), \( b = -12 \), and \( c = 11 \), the roots can be calculated, allowing us to continue solving for \( x \) by substituting the values back.

The discriminant is a key component of the quadratic formula that helps us determine the nature of the roots of a quadratic equation. The discriminant is given by the following expression:\[b^2 - 4ac\]It impacts the quadratic roots in the following ways:

• If the discriminant is greater than zero, the equation has two distinct real roots.
• If it equals zero, there is exactly one real root, known as a "repeated" or "double" root.
• If it is less than zero, the equation has no real roots, but instead, two complex roots.

In the given exercise, we calculate the discriminant for \( y^2 - 12y + 11 = 0 \) as \( 100 \), which is greater than zero. This tells us that the equation has two distinct real solutions. Thus, once the value of the discriminant confirms this, we can use the quadratic formula to clearly identify these roots.

Finding algebraic solutions for a quadratic equation involves not just computation but also interpretation of results in the context of the original problem. Once we have applied the quadratic formula to solve for \( y \) in the disguised quadratic equation \( y^2 - 12y + 11 = 0 \), we find two solutions: \( y = 11 \) and \( y = 1 \). These solutions are not the final answers, though, because we initially set \( y = x^2 \). We must substitute these values back to find the real roots of \( x \):

• For \( y = 11 \), since \( y = x^2 \), we have \( x^2 = 11 \), leading to \( x = \pm \sqrt{11} \).
• For \( y = 1 \), we have \( x^2 = 1 \), leading to \( x = \pm 1 \).

The complete set of solutions for the original quartic equation is thus \( x = \pm 1 \) and \( x = \pm \sqrt{11} \). This reveals the full solution set and closes the problem effectively, demonstrating how algebraic principles help us solve polynomial equations.