Problem 18 - Algebra 2 and Trigonometry

Question

Solve each proportion for the variable. \(\frac{x-2}{x}=\frac{3}{x+2}\)

Step-by-Step Solution

Answer

The solution is \(x = 4\).

Step 1: Cross Multiply

In a proportion like \(\frac{a}{b} = \frac{c}{d}\), you can cross multiply to get rid of fractions. Multiply \(a \times d\) and \(b \times c\). For our equation, multiply \((x-2)(x+2)\) and \(x \times 3\). The equation becomes \((x-2)(x+2) = 3x\).

Step 2: Expand the Left Side

Expand the left side of the equation using the distributive property. \((x-2)(x+2)\) is a difference of squares formula, \(a^2 - b^2\), resulting in \(x^2 - 4\). The equation is now \(x^2 - 4 = 3x\).

Step 3: Rearrange into a Standard Quadratic Equation

Rearrange the equation \(x^2 - 4 = 3x\) to get all terms on one side. Subtract \(3x\) from both sides: \(x^2 - 3x - 4 = 0\). This gives us a standard quadratic equation.

Step 4: Factor the Quadratic Equation

Determine two numbers that multiply to \(-4\) (the constant term) and add to \(-3\) (the coefficient of \(x\)). The numbers \(-4\) and \(1\) work because \(-4 \times 1 = -4\) and \(-4 + 1 = -3\). This allows us to factor as \((x - 4)(x + 1) = 0\).

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\). So, \(x-4=0\) or \(x+1=0\). Solving these gives \(x=4\) or \(x=-1\).

Step 6: Verify the Solutions

Substitute each solution back into the original proportion to ensure no division by zero or false equalities occur. For \(x=4\), \(\frac{4-2}{4} \overset{?}{=} \frac{3}{4+2}\), which simplifies to \(\frac{2}{4} = \frac{3}{6}\), both sides are \(\frac{1}{2}\), hence it holds. For \(x=-1\), \(\frac{-1-2}{-1}\) is not valid due to division by zero. Thus, \(x = -1\) is invalid.

Key Concepts

Cross Multiplication

Cross multiplication is an essential technique when solving proportions. A proportion is an equation where two ratios are set equal, such as \( \frac{a}{b} = \frac{c}{d} \). To solve for a variable, you multiply across the equal sign diagonally: \( a \times d = b \times c \). This method eliminates the fractions, simplifying the equation into a format that's often easier to handle.
For example, given the equation \( \frac{x-2}{x} = \frac{3}{x+2} \), cross multiplying results in \( (x-2)(x+2) = 3x \). After cross multiplying, you can solve for the variable by working with this new, simpler equation.

Quadratic Equation

A quadratic equation has the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. To solve a quadratic equation means finding the value(s) of \( x \) that make the equation true. These equations are called "quadratic" because the highest power of \( x \) is 2.
In our example, the cross-multiplied equation \( (x-2)(x+2) = 3x \) becomes \( x^2 - 4 = 3x \) after using the difference of squares. Rearranging gives \( x^2 - 3x - 4 = 0 \), which is a standard quadratic form. This form is the starting point for solving by various methods, including factoring, completing the square, or using the quadratic formula.

Factoring

Factoring is a method used to break down an expression into simpler parts, or factors, that when multiplied together give you the original expression. For quadratic equations, this allows you to set each factor to zero and solve for the unknown variable.
In the quadratic \( x^2 - 3x - 4 = 0 \), we can find two numbers that multiply to the constant term, \(-4\), and add to the linear coefficient, \(-3\). These numbers are \(-4\) and \(1\). Thus, we factor the quadratic as \((x - 4)(x + 1) = 0 \). Each of these factors represents a potential solution, which can be solved individually.

Difference of Squares

The difference of squares is a specific factoring pattern applied when dealing with certain quadratic expressions. It follows the formula \( a^2 - b^2 = (a + b)(a - b) \). This pattern is particularly handy because it allows you to quickly factor expressions that might otherwise seem complex.
In this exercise, the expression \((x - 2)(x + 2)\) fits the difference of squares pattern with \( a = x \) and \( b = 2 \), simplifying to \( x^2 - 4 \). Applying the difference of squares makes it straightforward to rewrite the expression without expanding everything the long way, saving both time and effort when solving equations.