Problem 24 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ p^{2}+\frac{p}{3}=\frac{1}{6} $$

Step-by-Step Solution

Answer

p = \frac{-1 + \sqrt{7}}{6} and p = \frac{-1 - \sqrt{7}}{6}

Step 1 - Rewrite the equation in standard form

First, rewrite the given equation in the standard form of a quadratic equation, which is: \[ax^2 + bx + c = 0\] Starting with: \[ p^2 + \frac{p}{3} = \frac{1}{6} \] Subtract \(\frac{1}{6}\) from both sides to set the equation to zero: \[ p^2 + \frac{p}{3} - \frac{1}{6} = 0 \].

Step 2 - Identify coefficients a, b, and c

Compare the equation \( p^2 + \frac{p}{3} - \frac{1}{6} = 0 \) with the standard form \( ax^2 + bx + c = 0 \). The coefficients are: \[ a = 1, \, b = \frac{1}{3}, \, c = -\frac{1}{6} \].

Step 3 - Write the quadratic formula

The quadratic formula is: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

Step 4 - Calculate the discriminant

Calculate the discriminant \( b^2 - 4ac \):\[ b^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \]\[ 4ac = 4 \cdot 1 \cdot \left( -\frac{1}{6} \right) = -\frac{4}{6} = -\frac{2}{3} \]\[ b^2 - 4ac = \frac{1}{9} - \left( -\frac{2}{3} \right) = \frac{1}{9} + \frac{6}{9} = \frac{7}{9} \].

Step 5 - Substitute values into the quadratic formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula: \[ p = \frac{-\left( \frac{1}{3} \right) \pm \sqrt{\frac{7}{9}}}{2 \cdot 1} \].

Step 6 - Simplify the expression

Simplify the expression under the square root and the overall formula: \[ p = \frac{-\frac{1}{3} \pm \frac{\sqrt{7}}{3}}{2} = \frac{-1 \pm \sqrt{7}}{6} \]. Therefore, the solutions are: \[ p = \frac{-1 + \sqrt{7}}{6} \quad \text{and} \quad p = \frac{-1 - \sqrt{7}}{6} \].

Key Concepts

solving quadratic equations

Solving quadratic equations involves finding the values of the variable that make the equation true. There are several methods to solve a quadratic equation, such as factoring, completing the square, and using the quadratic formula. The quadratic formula is particularly useful because it can solve any quadratic equation, regardless of whether the quadratic can be factored.

The quadratic formula is derived from the standard form of a quadratic equation, and it states:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. In this formula, \(a, b, \) and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\). The symbol ± indicates that there are usually two solutions, one by adding the square root and one by subtracting it. This formula will always give you the precise solutions for x when dealing with quadratic equations.

Here is a structured approach to using the quadratic formula:

Rewrite the equation in the standard form \(ax^2 + bx + c = 0\).

Identify the coefficients \(a, b,\) and \(c\).

Calculate the discriminant \(b^2 - 4ac\) to determine the nature and number of solutions.

Substitute \(a, b,\) and \(c\) into the quadratic formula to solve for \(x\).

Simplify the resulting expressions to find the solutions.

discriminant

The discriminant is a key part of the quadratic formula \(\sqrt{b^2 - 4ac}\). It helps determine the nature of the roots of a quadratic equation. Here's what different discriminant values mean:

If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real solutions.

If \(b^2 - 4ac = 0\), the quadratic equation has exactly one real solution (also known as a repeated or double root).

If \(b^2 - 4ac < 0\), the quadratic equation has two complex solutions, meaning no real solutions exist.

The discriminant essentially tells you how many times the parabola (graph of the quadratic equation) will intersect the x-axis. For example, in the provided exercise:

The equation was \(p^2 + \frac{p}{3} - \frac{1}{6} = 0\)

Calculating the discriminant involves the following steps:

Calculate \(b^2\). For \(b=\frac{1}{3}\), \(b^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\).

Compute \(4ac\), where \(a=1\) and \(c=-\frac{1}{6}\). So, \(4ac = 4 \times 1 \times -\frac{1}{6} = -\frac{2}{3}\).
Combine the results to find the discriminant: \(\frac{1}{9} - (-\frac{2}{3}) = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}\).

Finally, since \(\frac{7}{9} > 0\), the discriminant confirms two real solutions.

standard form of quadratic equation

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). This format allows us to use various algebraic methods to solve the equation effectively. In any quadratic equation, \(a, b,\) and \(c\) are constants, and \(a ≠ 0\). Every quadratic equation can be rearranged into this form if it's not already:

Move all terms to one side to set the equation to zero.

Combine like terms if necessary.

For instance, in the provided exercise:

The original equation was \(p^2 + \frac{p}{3} = \frac{1}{6}\).
To convert this into the standard form, we need to move \(\frac{1}{6}\) to the left side by subtracting it from both sides:
\(p^2 + \frac{p}{3} - \frac{1}{6} = 0\).

Now the equation fits the standard form where:
- \(a = 1\),
- \(b = \frac{1}{3}\),
- \(c = -\frac{1}{6}\).
This standardized format is crucial because it makes applying the quadratic formula, factoring, or completing the square a straightforward process. Always start solving a quadratic equation by rewriting it in the standard form.