Problem 59 - Introductory Algebra for College Students

Question

Will help you prepare for the material covered in the next section. In each exercise, evaluate $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ for the given values of \(a, b,\) and \(c .\) Where necessary, express answers in simplified radical form. \(a=9, b=-12, c=4\)

Step-by-Step Solution

Answer

The solution to the quadratic equation for the given values of a, b, and c is \(x = 2/3\) (double root).

Step 1: Substitution of values

Firstly, substitute the given values into the formula. The quadratic equation solver formula \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) becomes \(\frac{-(-12) \pm \sqrt{(-12)^{2}-4*9*4}}{2*9}\) after substitution.

Step 2: Calculate the discriminant

Then, calculate the value under the square root, often referred to as the discriminant. The expression \((-12)^{2}-4*9*4\) becomes \(144 - 144 = 0\). This reduces our formula to \(\frac{12 \pm \sqrt{0}}{18}\). Because the square root of zero is zero, this simplifies to \(\frac{12 \pm 0}{18}\).

Step 3: Calculate the solution(s) to the quadratic equation

Lastly, calculate the two solutions of the equation by using both the addition and the subtraction. The two solutions are \(\frac{12 + 0}{18} = \frac{12}{18}\) which simplifies to \(2/3\) and \(\frac{12 - 0}{18} = \frac{12}{18}\) which also simplifies to \(2/3\).

Key Concepts

Understanding the Discriminant

The discriminant is a crucial component of the quadratic formula used to determine the nature of the roots of a quadratic equation. It is represented by the expression under the square root in the quadratic formula: \(b^2 - 4ac\).
The discriminant can tell us a lot about the solutions to a quadratic equation:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it is zero, there is exactly one real root, often referred to as a repeated or double root.
• If the discriminant is negative, the equation has two complex roots.

In the scenario provided, the discriminant comes out to be zero (\(144 - 144 = 0\)), which leads to a single, repeated solution for the quadratic equation. Understanding and calculating the discriminant is essential as it helps us identify the type and number of solutions quickly.

Simplified Radical Form Explained

The simplified radical form is used to express numbers in a more concise and manageable way. When dealing with quadratic equations, this often involves simplifying the expression under the square root or the entirety of the solution.

• Simplifying involves breaking down the numbers under the square root and reducing them to their simplest form.
• For instance, if you have a perfect square under the square root, like \(\sqrt{0}\), it simplifies directly to 0.
• This process also applies to other cases, potentially involving square roots of non-perfect squares, which might not reduce as neatly as our example but can still be expressed more simply.

In the exercise example, the discriminant was zero, so the expression inside the square root was already as simple as possible: \(\sqrt{0} = 0\). This made solving the rest of the equation straightforward. Expressing quadratic solutions in simplified radical form provides clarity and ease of calculation.

Finding Quadratic Equation Solutions

Solving a quadratic equation typically results in discovering its roots, and the quadratic formula is one of the most efficient ways to find these solutions.
Given the formula \(-b \pm \sqrt{b^{2}-4ac})/2a\), the steps to finding the quadratic equation's solutions include:

• Substituting the values of \(a, b,\) and \(c\) into the formula.
• Calculating the discriminant, \(b^2 - 4ac\), to assess the number and nature of solutions.
• Using the plus-minus operator (\(\pm\)) to find two potential solutions when applicable.
• In our example, this results in \(\frac{12 \pm 0}{18}\), which simply resolves to \(\frac{2}{3}\).

Because the discriminant was zero, the plus-minus term here did not change the outcome, resulting in a single solution. Successfully solving quadratic equations involves recognizing these patterns and using the quadratic formula effectively.