Problem 6 - Basic Technical Mathematics with Calculus

Question

Find the roots of the given equations by inspection. $$\left(4 y^{2}+9\right)\left(25 y^{2}-10 y+1\right)=0$$

Step-by-Step Solution

Answer

The only real root is \(y = \frac{1}{5}\).

Step 1: Understand the Problem

We have a product of two polynomials, \((4y^2 + 9)\) and \((25y^2 - 10y + 1)\), that equals zero. To find the roots, we'll set each polynomial separately equal to zero.

Step 2: Solve First Polynomial

First, solve \(4y^2 + 9 = 0\). This equation implies that \(4y^2 = -9\). Since the square of a real number is always non-negative, \(y\) has no real solutions here.

Step 3: Solve Second Polynomial

Next, solve \(25y^2 - 10y + 1 = 0\). This is a quadratic equation that can potentially have real solutions. Use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 25\), \(b = -10\), and \(c = 1\).

Step 4: Apply Quadratic Formula

Substitute the values into the formula: \(y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}\). Simplify to \(y = \frac{10 \pm \sqrt{100 - 100}}{50}\).

Step 5: Simplify the Solutions

Since \(\sqrt{100 - 100} = \sqrt{0} = 0\), the solutions to the quadratic are \(y = \frac{10}{50} = \frac{1}{5}\). Therefore, \(y = \frac{1}{5}\) is the only real solution.

Key Concepts

Understanding Polynomials

Polynomials are mathematical expressions consisting of variables and coefficients. They are made up of terms, which may have different powers of the same variable. For instance, the expression \(4y^2 + 9\) is a polynomial because it includes a variable \(y\) raised to a power (in this case, the power is 2). This polynomial has just one term with the variable \(y^2\) and a constant term, 9.

Polynomials play a significant role in algebra because they form the basis of various functions and equations. When we are given a polynomial equation, such as \((25y^2 - 10y + 1) = 0\), our job is to find the values of \(y\), also known as the roots, that make the equation true. This involves solving the equation by setting it to zero, as shown in the exercise.

Solving polynomial equations often means using different techniques like factoring, inspection, or applying complex formulas depending on the degree and form of the polynomial.

Finding Real Solutions

In mathematics, particularly algebra, we often want to find the 'real solutions' to equations. Real solutions refer to the values that satisfy an equation using real numbers. In the context of our original exercise, the task was to find real solutions to polynomial equations.

In our step-by-step solution, we encountered two polynomial equations: \(4y^2 + 9 \) and \(25y^2 - 10y + 1\). For \(4y^2 + 9\), we quickly noticed that it had no real solutions. Why? This polynomial simplifies to an equation where \(y^2 = -\frac{9}{4}\), which isn’t possible with real numbers since squares of real numbers cannot be negative.

Meanwhile, \(25y^2 - 10y + 1\) was a different story. It had the potential for real solutions, and by using the quadratic formula, we were able to determine its real root: \(y = \frac{1}{5}\). Hence, always check the feasibility of having real solutions based on the mathematical properties of the equation when examining polynomials.

Applying the Quadratic Formula

The quadratic formula is a fundamental tool in algebra that helps us find the roots or solutions of quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Understanding and applying this formula was crucial in solving the second polynomial in our exercise:

1. Identify the coefficients \(a\), \(b\), and \(c\) from the polynomial. For \(25y^2 - 10y + 1 = 0\), we have \(a = 25\), \(b = -10\), and \(c = 1\).
2. Substitute these values into the quadratic formula to solve for \(y\): \(y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}\).
3. Simplify to find \(y = \frac{10 \pm \sqrt{0}}{50}\).

This step reveals the beautiful symmetry in math where the discriminant \(b^2 - 4ac\) dictates the nature of the solutions. Here, it resulted in a single real solution since the discriminant was zero, meaning the quadratic touched the x-axis at exactly one point: \(y = \frac{1}{5}\).

Mastering the quadratic formula makes solving quadratic equations straightforward and demonstrates the elegance of mathematical problem-solving.