Problem 54 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$y 2-3.6 y+2.03=0$$

Step-by-Step Solution

Answer

The solutions are \(y = 2.9\) and \(y = 0.7\).

Step 1: Identify the coefficients

The given equation is \(y^2 - 3.6y + 2.03 = 0\). First, identify the coefficients from this quadratic equation in the standard form \(ay^2 + by + c = 0\). Here, \(a = 1\), \(b = -3.6\), and \(c = 2.03\).

Step 2: Write the quadratic formula

The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula will be used to find the solutions for \(y\).

Step 3: Calculate the discriminant

The discriminant \(\Delta\) is given by \(b^2 - 4ac\). Substitute the values to find \(\Delta = (-3.6)^2 - 4 \cdot 1 \cdot 2.03\). Calculate \(\Delta = 12.96 - 8.12 = 4.84\).

Step 4: Compute the square root of the discriminant

Find the square root of the discriminant: \(\sqrt{4.84} = 2.2\).

Step 5: Apply the quadratic formula

Substitute \(b = -3.6\), \(\sqrt{\Delta} = 2.2\), and \(a = 1\) into the quadratic formula: \(y = \frac{-(-3.6) \pm 2.2}{2 \cdot 1}\). Simplify to get \(y = \frac{3.6 \pm 2.2}{2}\).

Step 6: Solve for the roots

Calculate the two values of \(y\):1. \(y_1 = \frac{3.6 + 2.2}{2} = \frac{5.8}{2} = 2.9\)2. \(y_2 = \frac{3.6 - 2.2}{2} = \frac{1.4}{2} = 0.7\)

Step 7: Write the solution

The solutions for the quadratic equation \(y^2 - 3.6y + 2.03 = 0\) are \(y = 2.9\) and \(y = 0.7\).

Key Concepts

Quadratic Equations

Quadratic equations are special types of polynomial equations characterized by their highest degree term being a square. In general, a quadratic equation takes the standard form:

• \[ ax^2 + bx + c = 0 \]

Here, \(a\), \(b\), and \(c\) are coefficients, with \(a eq 0\). This ensures that the equation is quadratic rather than linear. Quadratic equations frequently appear in real-world scenarios. They help describe physical properties like areas and projectile paths.
To solve a quadratic equation, there are several methods available:

• Factoring, when the quadratic is factorable.
• Completing the square, which rewrites the equation in a form that reveals the solutions.
• Using the quadratic formula, a versatile and powerful tool that we utilized in this exercise.

Understanding the form and tools to handle these equations is crucial for solving them effectively.

Discriminant

The discriminant is a key component in analyzing quadratic equations. It is derived from the quadratic formula:

• \[ ext{Discriminant} = riangle = b^2 - 4ac \]

The value of the discriminant provides information about the nature of the roots of the equation:

• If \(\triangle > 0\), the quadratic equation has two distinct real roots.
• If \(\triangle = 0\), there is exactly one real root. This is also known as a repeated or double root.
• If \(\triangle < 0\), there are no real roots; instead, the roots are complex or imaginary.

In our exercise, the discriminant was calculated as \(\triangle = 4.84\), which is greater than zero. This means the equation \(y^2 - 3.6y + 2.03 = 0\) has two distinct real roots.

Roots of Equations

Roots of quadratic equations are the solutions obtained that satisfy the equation, meaning when you substitute these solutions back into the equation, they hold true. Typically, for a quadratic equation, there are two roots. These can be real or complex numbers.
The quadratic formula provides a reliable method for finding these roots. With our given equation and after calculating the discriminant as described:

• \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

We used this formula to find the roots for \(y\). The discriminant helped us conclude that there are real roots, leading to the solutions \(y = 2.9\) and \(y = 0.7\). These values are the specific answers for the equation \(y^2 - 3.6y + 2.03 = 0\). They represent points where the quadratic curve intersects the x-axis when plotted.
Understanding the nature of the roots is essential for interpreting the solutions in practical contexts, providing insights into the equation's behavior.