Problem 80 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ p(p+1)=6 $$

Step-by-Step Solution

Answer

The solutions are \(p = -3\) and \(p = 2\).

Step 1: Expand the equation

Start by expanding the left side of the equation. The equation is given as: \[ p(p+1) = 6 \]Expanding, we get: \[ p^2 + p = 6 \]

Step 2: Rearrange the equation

Next, move all terms to one side of the equation to set it equal to zero:\[ p^2 + p - 6 = 0 \]

Step 3: Factor the quadratic equation

Now we need to factor the quadratic equation:\[ p^2 + p - 6 \]We look for two numbers that multiply to \(-6\) (the constant term) and add up to \(1\) (the coefficient of the linear term). These numbers are \(3\) and \(-2\):\[ (p + 3)(p - 2) = 0 \]

Step 4: Solve for p

Set each factor equal to zero and solve for \(p\):1. \(p + 3 = 0\) leads to \(p = -3\)2. \(p - 2 = 0\) leads to \(p = 2\)

Step 5: Verify solutions

Check the solutions by substituting them back into the original equation:For \(p = -3\):\[ p(p+1) = (-3)(-3+1) = (-3)(-2) = 6 \]For \(p = 2\):\[ p(p+1) = (2)(2+1) = (2)(3) = 6 \]Both values satisfy the original equation.

Key Concepts

Factoring

Factoring is an essential technique for solving quadratic equations like the one in this exercise. Here, the quadratic equation is transformed through factoring into a product of binomials. This process involves finding two numbers that fit two key conditions:

• Their product should equal the constant term from the quadratic equation.
• Their sum should equate to the coefficient of the linear term.

In our problem, we have the equation \( p^2 + p - 6 = 0 \). The constant term is \(-6\) and the coefficient of the linear term is \(+1\). The goal is to find two numbers, which in this case are \(+3\) and \(-2\), that multiply to \(-6\) and add to \(+1\). These values allow us to factor the quadratic expression into \((p + 3)(p - 2) = 0\). By factoring, we convert the quadratic equation into a simpler form that can be solved more straightforwardly. This results in solutions for \(p\) when each binomial expression is set to zero.

Expanding Expressions

Expanding expressions means transforming a condensed or factored equation into its complete form. This technique is useful when you start with a product of terms and need to write them as a full quadratic equation. In this specific exercise, we start with the equation \( p(p+1) = 6 \). To expand, you multiply \( p \) with both terms within the parentheses: \[ p \times p = p^2 \] and \[ p \times 1 = p \].So, the expansion results in the quadratic expression \( p^2 + p = 6 \). This step is crucial because it sets the stage for further transformations like rearranging and factoring.

Verifying Solutions

Verifying solutions is the final step in solving a quadratic equation. After finding potential solutions, it's crucial to substitute these back into the original equation to ensure they work.To do this, substitute each solution obtained from the factored equation back into the original equation \( p(p+1) = 6 \). This confirms if the left-hand side equals the right-hand side:1. For \( p = -3 \), - Substitute: \( (-3)((-3) + 1) = (-3)(-2) = 6 \).2. For \( p = 2 \), - Substitute: \( (2)(2 + 1) = (2)(3) = 6 \).Both substitutions show the solutions satisfy the original equation, confirming their correctness. Verifying is important because it ensures accuracy and helps identify any potential errors in earlier steps.