Problem 52 - Elementary and Intermediate Algebra

Question

Solve each proportion. $$ \frac{5 h}{14 h+3}=\frac{1}{h} $$

Step-by-Step Solution

Answer

The solutions are \( h = 3 \) and \( h = -0.2 \).

Step 1: Identify the Cross Products

Given the proportion \( \frac{5h}{14h+3} = \frac{1}{h} \), the first step is to use the cross product property of proportions, which states that the product of the means equals the product of the extremes. Set up the equation from the proportion: \( 5h \times h = 1 \times (14h + 3) \).

Step 2: Simplify Both Sides

Simplify the equation obtained from Step 1. The left side becomes \( 5h^2 \), and the right side distributes to give \( 14h + 3 \). So, we have: \( 5h^2 = 14h + 3 \).

Step 3: Form a Quadratic Equation

Rearrange the equation from Step 2 to form a standard quadratic equation: \( 5h^2 - 14h - 3 = 0 \).

Step 4: Use the Quadratic Formula

Since the equation \( 5h^2 - 14h - 3 = 0 \) is a quadratic equation, use the quadratic formula, \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = -14 \), and \( c = -3 \). Substitute these values into the formula.

Step 5: Calculate the Discriminant

Calculate the discriminant: \( b^2 - 4ac = (-14)^2 - 4 \times 5 \times (-3) = 196 + 60 = 256 \). Since the discriminant is positive, there will be two distinct real solutions.

Step 6: Find the Roots Using the Quadratic Formula

Substitute the values into the quadratic formula: \( h = \frac{-(-14) \pm \sqrt{256}}{2 \times 5} \). Simplifying gives \( h = \frac{14 \pm 16}{10} \).

Step 7: Compute the Solutions for 'h'

Calculate the two possible solutions: \( h = \frac{14 + 16}{10} = 3 \) and \( h = \frac{14 - 16}{10} = -0.2 \).

Step 8: Verify Solutions

Check the solutions by substituting back into the original proportion to verify. Substitute \( h = 3 \) and \( h = -0.2 \) into \( \frac{5h}{14h+3} = \frac{1}{h} \) to check for consistency.

Key Concepts

Cross Product Property

The cross product property is a fundamental rule in solving proportions. It states that for any proportion \( \frac{a}{b} = \frac{c}{d} \), the product of the means (\(b \times c\)) is equal to the product of the extremes (\(a \times d\)). In simpler terms, when you have a proportion, you can cross multiply to help solve for an unknown variable.

Let's use our example, \( \frac{5h}{14h+3} = \frac{1}{h} \). Using the cross product property, we multiply the numerator of the first ratio by the denominator of the second ratio and set it equal to the opposite cross multiplication:

• \(5h \times h = 5h^2\)
• \(1 \times (14h + 3) = 14h + 3\)

This gives us the equation \(5h^2 = 14h + 3\), which is crucial for setting up a quadratic equation.

Quadratic Equation

A quadratic equation is any polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). It usually results from expressions involving squares, like our equation \(5h^2 - 14h - 3 = 0\), which we acquired from the cross product property.

Quadratic equations are powerful because they describe a wide array of phenomena in mathematics and science. To solve them, we sometimes need to rearrange terms to ensure we have it in the standard form. In our case, starting from \(5h^2 = 14h + 3\) and moving terms to one side results in \(5h^2 - 14h - 3 = 0\), a classic setup ripe for solving using formulaic methods.

Quadratic Formula

The quadratic formula is a universal method for finding solutions to quadratic equations, even when they cannot be easily factored. It is expressed as: \[h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is derived from completing the square in a quadratic equation and is very efficient. For the equation \(5h^2 - 14h - 3 = 0\), we identify values of \(a = 5\), \(b = -14\), and \(c = -3\) from the standard form.

By substituting these into the quadratic formula, we can solve for \(h\). The presence of the \(\pm\) indicates that there could be two solutions, which are influenced by the square root component, known as the discriminant. We observed in this case that solving yields two potential solutions: \(h = 3\) and \(h = -0.2\).

Discriminant

The discriminant is a key component of the quadratic formula, under the square root, represented as \(b^2 - 4ac\). It helps determine the nature of the roots without solving the equation completely. Based on its value, we understand:

• If it is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (roots are repeated).
• If it is negative, there are no real solutions, but two complex solutions.

In our equation \(5h^2 - 14h - 3 = 0\), computing the discriminant gives us \(256\), which is positive. This indicates that there are two distinct real solutions, aligning with our results of \(h = 3\) and \(h = -0.2\). Understanding the discriminant gives insight into the potential outcomes of solving quadratic equations.