Problem 60 - Algebra and Trigonometry

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. $$\left(y-\frac{10}{y}\right)^{2}+6\left(y-\frac{10}{y}\right)-27=0$$

Step-by-Step Solution

Answer

The solution to the given equation involves simplifying it through a substitution, solving the resulting quadratic equation, and reverting the substitution to find possible values of \( y \).

Step 1: Formulate a Substitution

Substitute \( u = y - \frac{10}{y} \) in the equation. This is done to simplify the equation and form a quadratic equation that can be easily solved.

Step 2: Substitute in the Equation

The equation transforms as: \( u^{2} + 6u - 27 = 0 \). Now it is simpler to solve.

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \( u^{2} + 6u - 27 = 0 \) by factoring it, or by using the quadratic formula if it cannot be factored.

Step 4: Substitute the values of u into the original substitution equation

Solving the quadratic equation gives the values for \( u \). Substitute these back in equation \( u = y - \frac{10}{y} \) to get the possible values of \( y \) in the original problem.

Step 5: Solve the equation for y

Solving the equation \( u = y - \frac{10}{y} \) will give us the final solutions for \( y \) in the original equation. If the equation cannot be solved directly, use the quadratic formula if it leads to a quadratic equation.

Key Concepts

Quadratic Equations

Quadratic equations are a vital part of algebra. These are equations that can be represented by the standard form \( ax^2 + bx + c = 0 \). In this form, \( a \), \( b \), and \( c \) are coefficients, where \( a \) is not equal to zero. The solution to a quadratic equation often involves finding the values of \( x \) that satisfy the equation, also known as the roots of the equation.

• Quadratics can have two real and distinct roots, two real and identical roots, or two complex roots.
• The solutions can be found using various methods, such as factoring, completing the square, or using the quadratic formula.

In our exercise, the challenge is to solve a transformed quadratic equation, \( u^2 + 6u - 27 = 0 \). Whether via substitution or direct solving, understanding the nature of quadratic equations helps to simplify complex algebraic problems.

Algebraic Substitution

Algebraic substitution is a powerful technique to simplify complex equations, making them easier to solve. This involves replacing a part of the equation with a single variable. The key idea is to transform an unwieldy expression into a more standard form that is easier to handle numerically or algebraically.

• Identify a repeat element or a complex expression in the equation.
• Substitute that expression with a new variable.
• Solve the resulting simpler equation.
• Substitute back the values to get the results for the original variable.

In the given problem, we used substitution to transform \( y - \frac{10}{y} \) into \( u \). This crucial step turned the original complex equation into a simplistic quadratic form, allowing for easier solving and clear interpretation.

Factoring Quadratics

Factoring is one of the simplest and most reliable methods to solve quadratic equations when they can be factored. It involves expressing the quadratic equation as a product of two binomials. The equation then looks like \((x - m)(x - n) = 0\), where \( m \) and \( n \) are the solutions or roots of the quadratic equation.

• Ensure the quadratic is in the standard form \( ax^2 + bx + c = 0 \).
• Look for two numbers that multiply to \( ac \) and add to \( b \).
• Rewrite the middle term using these two numbers and factor by grouping.

For our equation \( u^2 + 6u - 27 = 0 \), factoring, if possible, provides a straightforward path to solving for \( u \). If factoring is not feasible, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) serves as a reliable alternative.