91Ó°ÊÓ

Factorization is a technique used to simplify algebraic expressions, particularly polynomials, by converting them into a product of simpler expressions. This can make solving equations more manageable. In our solution, we applied factorization to a quadratic equation to find its roots.

When we reached the expression \(y^4 - 2y^2 + 1 = 0\), we recognized it as a perfect square trinomial. Perfect squares are special because they can be expressed as the square of a binomial. For the expression \(y^4 - 2y^2 + 1\), we factored it as \((y^2 - 1)^2 = 0\).

Here's a simple way to understand this: a perfect square trinomial like \(a^2 - 2ab + b^2\) can be rewritten as \((a-b)^2\). In our problem, \(a\) is \(y^2\) and \(b\) is 1, leading to the factorization \((y^2 - 1)^2\). Through factorization, we reduced the problem to solving \(y^2 - 1 = 0\), which is much simpler.

The substitution method is a powerful algebraic tool that replaces one variable or expression with another value or expression to simplify an equation. It often helps when dealing with complex equations by reducing them to a simpler form.

In our problem, we used substitution to transform the equation \((x^2 - 1)^2 + (x^2 - 1)^{-2} = 2\). By setting \(y = x^2 - 1\), we simplified the equation to \(y^2 + y^{-2} = 2\). This conversion makes it easier to manage because we're dealing with just one variable, \(y\), instead of a more complex expression in \(x\).

Substitution often involves:

• Identifying a complex or repetitive part of an equation.
• Assigning a new variable to this part (like \(y = x^2 - 1\)).
• Rewriting the original equation with the new variable to simplify the expression.

This method is especially useful in solving algebraic equations involving multiple terms or exponents.

Quadratic equations are polynomial equations of degree two, usually presented in the form \(ax^2 + bx + c = 0\). Solving quadratic equations is a common problem in algebra, and there are several methods to find the solutions (or roots) of these equations.

In the problem at hand, after substitution and manipulation, we arrived at a quadratic equation in terms of \(y^2\): \(y^2 - 1 = 0\). This is a simpler form of a quadratic equation, where the variable solves to \(y^2 = 1\).

Quadratic equations can be solved using various methods:

• **Factoring:** Expressing the quadratic as a product of two binomials, if possible.
• **Completing the Square:** Rewriting the quadratic in a form that reveals the solution directly.
• **Quadratic Formula:** \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides a direct way to find the roots.

In our exercise, since \(y^2 - 1\) is a simple polynomial, the factorization method was quick and effective, resulting in solutions \(y = \pm1\). This demonstrates the versatility of quadratic solving methods.