Problem 34 - Elementary and Intermediate Algebra

Question

Solve each equation and check the result. If an equation has no solution, so indicate. $$ \frac{7}{p}-p=-6 $$

Step-by-Step Solution

Answer

The solutions are \( p = 7 \) and \( p = -1 \). Both satisfy the original equation.

Step 1: Eliminate Fractions

The given equation is \( \frac{7}{p} - p = -6 \). To eliminate the fraction, multiply every term of the equation by \( p \) to get rid of the denominator. This gives us: \[ 7 - p^2 = -6p \].

Step 2: Rearrange the Equation

Bring all the terms to one side to set the equation to zero, like a standard quadratic equation. Rearranging the terms gives us: \[ p^2 - 6p - 7 = 0 \].

Step 3: Solve the Quadratic Equation

The quadratic equation \( p^2 - 6p - 7 = 0 \) can be solved using the quadratic formula: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1 \), \( b = -6 \), and \( c = -7 \). Calculate the discriminant: \[ b^2 - 4ac = (-6)^2 - 4 \times 1 \times (-7) = 36 + 28 = 64 \].The roots are: \[ p = \frac{6 \pm \sqrt{64}}{2} = \frac{6 \pm 8}{2} \].This gives two solutions: \( p = 7 \) and \( p = -1 \).

Step 4: Verify the Solutions

Substitute \( p = 7 \) back into the original equation: \( \frac{7}{7} - 7 = 1 - 7 = -6 \), which is correct.Substitute \( p = -1 \) back into the original equation: \( \frac{7}{-1} - (-1) = -7 + 1 = -6 \), which is correct.

Key Concepts

Solving Equations

Solving equations is essentially about finding the value or values of a variable that make the equation true. In the problem at hand, we started with the equation \( \frac{7}{p} - p = -6 \). To simplify things and eliminate fractions, we must manipulate the equation. This is done by multiplying every term by the variable's denominator. Here, that means multiplying every term by \( p \).

This operation results in the new equation \( 7 - p^2 = -6p \). Now we've arrived at a manageable form without fractions, and the next step is to further transform it into something more familiar: a quadratic equation. We'll do this by moving all terms to one side, resulting in \( p^2 - 6p - 7 = 0 \).

• Start with the original equation.
• Eliminate fractions by multiplying all terms by the denominator.
• Rearrange to form a quadratic equation.

Through these steps, you set up the equation ready to solve using methods suitable for quadratic equations.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). In our situation, \( a = 1 \), \( b = -6 \), and \( c = -7 \). The values correspond to the coefficients in \( p^2 - 6p - 7 = 0 \).

The formula itself is: \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Using it requires calculating the discriminant: \( b^2 - 4ac \). This determines the nature and number of solutions:

• If the discriminant is positive, there will be two distinct real solutions.
• If it's zero, you'll get one real solution.
• If negative, the solutions are complex or imaginary.

For our equation, \( (-6)^2 - 4 \times 1 \times (-7) = 64 \), which is positive, means two real solutions. Plugging into the formula gives \( p = 7 \) and \( p = -1 \). Using the quadratic formula can reveal solutions that aren't immediately obvious and helps handle any quadratic equation efficiently.

Checking Solutions

It's crucial to check your solutions in the original equation to ensure they are correct. Sometimes, the modification to get to a quadratic equation can introduce extraneous solutions or errors.

For the equation \( \frac{7}{p} - p = -6 \), let's verify the solutions we found: \( p = 7 \) and \( p = -1 \).

Substitute \( p = 7 \) back into the original equation:

• Calculate \( \frac{7}{7} - 7 = 1 - 7 = -6 \).

This confirms \( p = 7 \) is valid.

Next, substitute \( p = -1 \):

• Calculate \( \frac{7}{-1} - (-1) = -7 + 1 = -6 \).

This confirms \( p = -1 \) is also valid.

By substituting and ensuring both sides match, you confirm the solutions accurately satisfy the original equation. This step is crucial for ensuring that no mistakes were made in solving the problem.