Problem 39 - Precalculus: Functions and Graphs

Question

Solve the equation. \(36 x^{-4}-13 x^{-2}+1=0\)

Step-by-Step Solution

Answer

The solutions are \( x = \pm 2 \) and \( x = \pm 3 \).

Step 1: Recognize the Substitution

Notice that the equation is in the form of a quadratic expression in terms of the variable substitution. Let \( y = x^{-2} \). The equation \( 36x^{-4} - 13x^{-2} + 1 = 0 \) becomes \( 36y^2 - 13y + 1 = 0 \).

Step 2: Identify Coefficients for the Quadratic Equation

With the equation in the form \( 36y^2 - 13y + 1 = 0 \), identify the coefficients: \( a = 36 \), \( b = -13 \), and \( c = 1 \).

Step 3: Apply the Quadratic Formula

Use the quadratic formula to solve for \( y \): \( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \). Substitute \( a = 36 \), \( b = -13 \), \( c = 1 \) into the formula.

Step 4: Calculate the Discriminant

Calculate the discriminant: \( b^2 - 4ac \). Here, \((-13)^2 - 4 \times 36 \times 1 = 169 - 144 = 25\).

Step 5: Solve for y

Substitute the discriminant back into the quadratic formula: \( y = \frac{{13 \pm 5}}{72} \). Solve: \( y = \frac{18}{72} = \frac{1}{4} \) and \( y = \frac{8}{72} = \frac{1}{9} \).

Step 6: Reverse the Substitution

Recall that \( y = x^{-2} \). Thus, \( x^{-2} = \frac{1}{4} \) and \( x^{-2} = \frac{1}{9} \).

Step 7: Solve for x

Solving for \( x \) from \( x^{-2} = \frac{1}{4} \), we have \( x = \pm 2 \). Solving for \( x^{-2} = \frac{1}{9} \), we have \( x = \pm 3 \).

Step 8: Verify Solutions

Check each solution in the original equation to verify correctness. The solutions \( x = \pm 2 \) and \( x = \pm 3 \) satisfy the equation.

Key Concepts

Substitution Method

In solving complex equations, substitution can be a powerful tool. This usually involves simplifying a complicated expression by replacing part of it with a new variable. In our exercise, the equation was initially in terms of \( x^{-4} \) and \( x^{-2} \). By substituting \( y = x^{-2} \), the complicated equation was transformed into a simpler quadratic equation in terms of \( y \). This makes it easier to solve the equation using known techniques. The key is to identify a substitution that reduces the equation to a simpler form.

Substitution helps in converting an equation in an unfamiliar form into something more recognizable and manageable, like turning a higher degree equation into a quadratic equation. Once the equation is solved with respect to the new variable, the original variable can be found using the reverse substitution.

Quadratic Formula

The quadratic formula is a universal method for finding solutions to a quadratic equation. Any equation of the form \( ax^2 + bx + c = 0 \) can be solved using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a \), \( b \), and \( c \) are coefficients in the quadratic equation. In our specific case, the coefficients were \( a = 36 \), \( b = -13 \), and \( c = 1 \).

The quadratic formula will provide two solutions, one for the '+' and one for the '-' in the \( \pm \) symbol. These solutions, \( y \), correspond to the roots of the quadratic equation. The formula is reliable and works every time, as long as the discriminant is non-negative, giving us real solutions.

Discriminant

The discriminant is a part of the quadratic formula crucial in determining the nature of the roots of the quadratic equation. It is represented by \( b^2 - 4ac \). In our exercise, the discriminant calculated was \( 25 \). The value of the discriminant reveals the type of solutions:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root (a repeated or double root).
• If the discriminant is negative, the equation has two complex roots (conjugates of each other).

In this case, a positive discriminant means that our quadratic equation has two distinct solutions. This guided us to solve the quadratic equation using the quadratic formula.

Verification of Solutions

After deriving potential solutions from solving an equation, it is vital to ensure that the solutions are valid. Verification involves substituting the solutions back into the original equation to check if they satisfy it.

For our problem, after finding \( x = \pm 2 \) and \( x = \pm 3 \), each value was plugged back into the original equation \( 36x^{-4} - 13x^{-2} + 1 = 0 \). Valid solutions should make the equation equal to zero. Verification is an essential step to confirm the correctness of the solutions and identify any calculation errors made during the solving process. This step gives confidence that the solutions correctly solve the equation as intended.