Problem 42 - Algebra and Trigonometry

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. $$x^{4}-13 x^{2}+36=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x=2, -2, 3, -3\).

Step 1: Substitution

Let \(x^{2}\) be \(y\), such that the equation becomes: \(y^{2}-13y+36=0\).

Step 2: Solving the Simplified Equation.

Solving for \(y\) in the equation \(y^{2}-13y+36=0\) can be factored to \((y-4)(y-9)=0\). Setting each factor equal to zero gives \(y=4\) and \(y=9\).

Step 3: Substitute Back

Substitute back \(y\) with \(x^{2}\) to get \(x^{2}=4\) and \(x^{2}=9\). Solving the equation \(x^{2}=4\) gives \(x=2, -2\) and solving the equation \(x^{2}=9\) gives \(x=3, -3\).

Key Concepts

Factoring Quadratic Equations

The first step in solving the given polynomial equation involves recognizing it in a quadratic form. Quadratic equations are typically written as \(ax^2 + bx + c = 0\). In this case, we make a substitution to fit this format. By letting \(y = x^2\), the original equation \(x^4 - 13x^2 + 36 = 0\) transforms into \(y^2 - 13y + 36 = 0\).

The goal in factoring is to break down the quadratic into a product of simpler expressions. Look for two numbers that multiply to the constant term (36) and add to the middle coefficient (-13). These numbers are -4 and -9. Thus, the factored form is \((y - 4)(y - 9) = 0\).

• Identifying the right substitution simplifies the problem.
• Search for two numbers multiplying to the constant term while also summing to the middle coefficient for factoring.
• Factorization leads to simpler equations that can be easily solved.

Solving Polynomial Equations

After factoring the quadratic equation into \((y - 4)(y - 9) = 0\), we take each factor individually to solve for \(y\).

Set each factor equal to zero:

• \(y - 4 = 0\) leads to \(y = 4\).
• \(y - 9 = 0\) leads to \(y = 9\).

These solutions correspond to the possible values of \(y\) for the transformed equation.

This process, called the zero product property, allows breaking down polynomials into manageable components. Each component, when set to zero, reveals possible solutions. This method is critical in polynomial equations as it facilitates finding solutions by reducing it to simple linear solutions.

Algebraic Equations Solutions

Once the possible solutions for \(y\) are found, the next step is returning to the original variable. Substitute back \(y\) with \(x^2\):

• From \(y = 4\), substitute to get \(x^2 = 4\). Solving gives \(x = 2\) or \(x = -2\).
• From \(y = 9\), substitute to get \(x^2 = 9\). Solving gives \(x = 3\) or \(x = -3\).

The substitution method allows us to handle complex algebraic expressions by simplifying into well-known forms.

Returning to the initial variable is crucial to ensure that all steps align with the problem's requirements. Evaluating at the original variable completes the solution process, confirming the roots of the initial polynomial equation.