Problem 48 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ \left(\frac{x}{x-2}\right)^{2}-\frac{2 x}{x-2}-15=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{5}{2} \) and \( x = \frac{3}{2} \).

Step 1: Identify the substitution variable

To simplify the equation, observe that the term \( \frac{x}{x-2} \) can be substituted. Let's assume \( y = \frac{x}{x-2} \). This will transform the given equation into a simpler quadratic form.

Step 2: Substitute into the equation

Substitute \( y = \frac{x}{x-2} \) into the equation \( \left(\frac{x}{x-2}\right)^{2}-\frac{2x}{x-2}-15=0 \). This gives \( y^2 - 2y - 15 = 0 \).

Step 3: Solve the quadratic equation

The equation \( y^2 - 2y - 15 = 0 \) is a standard quadratic equation. To solve it, we can factor it as \( (y - 5)(y + 3) = 0 \). Thus, the solutions are \( y = 5 \) and \( y = -3 \).

Step 4: Back-substitute and solve for x

With \( y = 5 \), we have \( \frac{x}{x-2} = 5 \). Solving for \( x \), cross-multiply to get \( x = 5(x-2) \) which simplifies to \( x - 5x = -10 \) or \( -4x = -10 \). Thus, \( x = \frac{5}{2} \).With \( y = -3 \), we have \( \frac{x}{x-2} = -3 \). Cross-multiplying gives \( x = -3(x-2) \). This simplifies to \( x + 3x = 6 \) or \( 4x = 6 \). Thus, \( x = \frac{3}{2} \).

Step 5: Verify the solutions

Check both values of \( x \) to ensure they do not make the original denominator zero. \( x = \frac{5}{2} \) leads to a denominator of \( \frac{1}{2} \) and \( x = \frac{3}{2} \) leads to \( -\frac{1}{2} \). Both values are valid as they do not make the denominator zero.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. Solving a quadratic equation can be achieved through various methods such as factoring, completing the square, or using the quadratic formula. The solution to a quadratic equation are the values of \( x \) that satisfy the equation. These solutions are also known as the roots of the equation.

To factor a quadratic equation like \( y^2 - 2y - 15 = 0 \) from our original problem, we look for two numbers that multiply to \(-15\) (the constant term) and add to \(-2\) (the coefficient of \( y \)). In this case, \(-5\) and \(-3\) fit the requirement, allowing us to factor as \((y - 5)(y + 3) = 0\). The roots are \( y = 5 \) and \( y = -3 \).

It's important to understand the way quadratic equations are structured because recognizing their form helps us pick an efficient solution method and confidently apply it to other similar problems.

Substitution Method

The substitution method is a powerful technique used to simplify complex algebraic equations, which often makes them easier to solve. This method involves replacing a complicated expression with a simpler variable, solving the equation with respect to that variable, and then substituting back to find the desired values of the original variable.

In our problem, we noticed that the expression \( \frac{x}{x-2} \) appears twice in a more complicated equation. By setting \( y = \frac{x}{x-2} \), we essentially reduce the complexity of the equation, transforming it into the more approachable quadratic form \( y^2 - 2y - 15 = 0 \).

After solving this equation for \( y \), we get \( y = 5 \) and \( y = -3 \). The next step is to back-substitute these values into \( y = \frac{x}{x-2} \) to solve for \( x \), leading us to find the solutions \( x = \frac{5}{2} \) and \( x = \frac{3}{2} \). This technique is particularly useful in situations where an equation contains repeating expressions that can be substituted for simplification.

Cross-Multiplication

Cross-multiplication is a technique mainly used to solve equations involving fractions. It consists of multiplying the numerator of one fraction by the denominator of the other, thereby eliminating the fractions altogether.

In our equation, once we have substituted \( y \) back with \( \frac{x}{x-2} \) and found \( y = 5 \) and \( y = -3 \), we solve for \( x \) by setting \( \frac{x}{x-2} = y \). This turns into the equations \( x = 5(x-2) \) and \( x = -3(x-2) \). To solve for \( x \), we cross-multiply: this means multiplying \( x \) with \(-2\) or \(2\) (denominator) and \( y \) with \( x-2 \). This simplifies the expressions and allows us to isolate and solve for \( x \).

Cross-multiplication is very effective in simplifying fractional equations as it eliminates the fractions, leaving behind a simple linear equation to solve. Just remember that this method works best when an equation is set up as a proportion, meaning one fraction equals another.