Problem 67 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{t}{10}=\frac{10}{t} $$

Step-by-Step Solution

Answer

The solutions are \( t = 10 \) and \( t = -10 \).

Step 1: Understanding the Proportion

The problem provides the equation \( \frac{t}{10} = \frac{10}{t} \). This is a proportion, meaning two ratios are equal. Our goal is to find the value of \( t \) that makes this equation true.

Step 2: Cross-Multiply

To solve the proportion, we can cross-multiply: multiply the numerator of the first fraction by the denominator of the second fraction and set it equal to the product of the denominator of the first fraction and the numerator of the second fraction. This gives us \( t \cdot t = 10 \cdot 10 \).

Step 3: Simplify the Equation

Simplifying the result of cross-multiplication, we get \( t^2 = 100 \). This is a quadratic equation where \( t \) is squared.

Step 4: Take the Square Root

To solve for \( t \), take the square root of both sides of the equation: \( t = \sqrt{100} \). This gives two possible solutions: \( t = 10 \) and \( t = -10 \), since both squared give 100.

Step 5: Verify the Solutions

Check the original equation with both potential solutions. For \( t = 10 \), the equation becomes \( \frac{10}{10} = \frac{10}{10} \), which is true. For \( t = -10 \), the equation becomes \( \frac{-10}{10} = \frac{10}{-10} \), which simplifies to \( -1 = -1 \), and is also true.

Key Concepts

Cross-Multiplication

Cross-multiplication is a powerful and straightforward technique used to solve proportions, which are equations expressing that two ratios are equal. When you have a proportion like \( \frac{a}{b} = \frac{c}{d} \), cross-multiplication involves multiplying the numerator of the first ratio (\( a \)) by the denominator of the second ratio (\( d \)), and setting it equal to the product of the denominator of the first ratio (\( b \)) and the numerator of the second ratio (\( c \)). This can be formally written as:

• \( a \cdot d = b \cdot c \)

This method helps in transforming a proportion into a simpler equation that is often easier to solve. By applying cross-multiplication to the given exercise \( \frac{t}{10} = \frac{10}{t} \), we derive \( t \cdot t = 10 \cdot 10 \), leading to \( t^2 = 100 \). This is the equation we will solve using further algebraic methods.

Quadratic Equation

A quadratic equation is a type of polynomial equation that involves a variable being squared (raised to the power of 2) and can be written in the standard form \( ax^2 + bx + c = 0 \). In the exercise, we obtained the quadratic equation \( t^2 = 100 \) from cross-multiplying the original proportion. Here, \( a = 1 \), \( b = 0 \), and \( c = -100 \). To solve this equation for \( t \), we seek values that satisfy the equation when substituted back into it. Quadratic equations can be solved using various methods including:

• Factoring (when applicable)
• Completing the square
• Using the quadratic formula
• Or, like in this case, simply taking the square root of both sides when it's of the form \( t^2 = k \)

Since our equation is simple and does not include a linear term (\( bx \)), taking the square root of both sides directly gives the solution.

Square Roots

The concept of square roots involves finding a number which, when multiplied by itself, gives the original number. This operation is the inverse of squaring a number. In the quadratic equation from the exercise, \( t^2 = 100 \), we take the square root of both sides to find \( t \). The square root of 100 is 10, but it is important to remember that both positive and negative values could be a solution. Thus, \( t = 10 \) and \( t = -10 \) are both valid solutions because:

• 10 squared is 100 \((10 \times 10 = 100)\)
• -10 squared is also 100 \((-10 \times -10 = 100)\)

Checking both potential solutions in the original proportion confirms their validity, as both satisfy the given equation. This showcases the importance of considering both potential square root solutions in quadratic equations.