Problem 50 - Elementary and Intermediate Algebra

Question

Solve each proportion. $$ \frac{s}{s-5}=\frac{s+5}{24} $$

Step-by-Step Solution

Answer

The solutions are \(s = 25\) and \(s = -1\).

Step 1: Understand the Proportion

We start with the given proportion \( \frac{s}{s-5} = \frac{s+5}{24} \). Our goal is to solve for the variable \( s \).

Step 2: Cross-Multiply

To eliminate the fractions, cross-multiply the terms. This means multiplying across the equal sign: \[ s \cdot 24 = (s - 5) \cdot (s + 5) \] This results in the equation, \[ 24s = (s - 5)(s + 5) \]

Step 3: Apply the Difference of Squares

Recognize that the right side is a difference of squares, \((s - 5)(s + 5)\), which equals \(s^2 - 25\). Thus the equation becomes: \[ 24s = s^2 - 25 \]

Step 4: Rearrange the Equation

Rearrange the terms to one side of the equation to set it to zero and form a quadratic equation: \[ s^2 - 24s - 25 = 0 \]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \( s^2 - 24s - 25 = 0 \) using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -24 \), and \( c = -25 \).Calculate the discriminant: \[ b^2 - 4ac = (-24)^2 - 4 \times 1 \times (-25) = 576 + 100 = 676 \]Since the discriminant is a perfect square, compute the solutions: \[ s = \frac{24 \pm \sqrt{676}}{2} = \frac{24 \pm 26}{2} \]The two potential solutions are:\( s_1 = \frac{50}{2} = 25 \) and \( s_2 = \frac{-2}{2} = -1 \).

Step 6: Verify the Solutions

Check both solutions to ensure they don't cause any denominators to become zero, which would invalidate the solution. Plug back into the original proportion:- For \( s = 25 \): \( \frac{25}{25-5} = \frac{25+5}{24} \rightarrow \frac{25}{20} = \frac{30}{24} \rightarrow \frac{5}{4} = \frac{5}{4} \), which holds true.- For \( s = -1 \): \( \frac{-1}{-1-5} = \frac{-1+5}{24} \rightarrow \frac{-1}{-6} = \frac{4}{24} \rightarrow \frac{1}{6} = \frac{1}{6} \), which also holds true.

Key Concepts

Difference of Squares

The difference of squares is a special algebraic identity that describes how two perfect squares can be subtracted. It is written in the form: \( (a - b)(a + b) = a^2 - b^2 \). This formula can be really handy in simplifying expressions and solving equations involving squares. In our exercise, we noticed that \((s-5)(s+5)\) fits this form perfectly, where \( a = s \) and \( b = 5 \).

• Recognizing the difference of squares allows us to replace the expression with \( s^2 - 25 \), which simplifies calculations considerably.
• This expression reveals a powerful shortcut for squaring differences and sums, often used in factoring and algebraic manipulation.

Learning to spot these patterns helps you approach seemingly complex problems with strategies that simplify and streamline the solution process. The more you practice identifying these patterns, the easier it will become to solve problems efficiently.

Cross-Multiplication

Cross-multiplication is an essential arithmetic technique often used to solve equations involving proportions. Proportions are statements that two ratios are equal, such as \( \frac{a}{b} = \frac{c}{d} \). To solve for one of the unknowns, you can eliminate the fractions by multiplying each side by the denominator of the opposite side.

• For the proportion in our exercise, \( \frac{s}{s-5} = \frac{s+5}{24} \), cross-multiplying involves multiplying \( s \) and 24, and \( (s-5) \) and \( (s+5) \). This results in the equation: \( 24s = (s-5)(s+5) \).
• It helps to "clear" the fractions and reduce the equation to a simpler form, making it easier to solve.

Cross-multiplication you will find extremely useful elsewhere in mathematics too. It lays a foundation for solving more complex algebraic equations and even for understanding functions and transformations.

Quadratic Formula

When faced with a quadratic equation such as \( ax^2 + bx + c = 0 \), the quadratic formula is a reliable method to find its solutions. The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula works for any quadratic equation, helping you uncover the two possible values for the variable. In our exercise, we used this formula to solve \( s^2 - 24s - 25 = 0 \).

• The coefficients were \( a = 1 \), \( b = -24 \), and \( c = -25 \). Calculating the discriminant, \( b^2 - 4ac \), helped determine the nature of the solutions (real and distinct, in this particular case, since we ended up with a positive discriminant).
• The solutions found were \( s = 25 \) and \( s = -1 \). These solutions were then verified to ensure they satisfied the original proportion.

The quadratic formula is a vital tool for unlocking solutions to complex quadratic problems, providing a structured pathway from setup to solution. Understanding its derivation and application will immensely benefit your problem-solving toolbox.