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 Equation Structure

The given equation is a product of two expressions: \((4y^2 + 9)(25y^2 - 10y + 1) = 0\). This indicates if either of these expressions equals zero, then the entire equation equals zero.

Step 2: Solve the First Expression

For \(4y^2 + 9 = 0\) to be true, \(4y^2 = -9\). This implies \(y^2 = \frac{-9}{4}\). A square of a real number can't be negative, so there are no real solutions from this expression.

Step 3: Solve the Second Expression

For \(25y^2 - 10y + 1 = 0\), try to factor it directly. Notice this is a perfect square: \((5y - 1)^2 = 0\). This implies \(5y - 1 = 0\).

Step 4: Find the Root from Factored Expression

Solving the equation \(5y - 1 = 0\) gives \(5y = 1\), leading to \(y = \frac{1}{5}\).

Step 5: Conclude with Real Solutions

Given that there are no real roots from the first expression and \(y = \frac{1}{5}\) from the second, the only real root of the equation is \(y = \frac{1}{5}\).

Key Concepts

Roots of Equations

Roots of an equation are the values that make the equation true or, put simply, the values of the variable that satisfy the equation. In a polynomial equation, these roots are the solutions that result when the polynomial is set to zero and solved. In our exercise, the equation

• \((4y^2 + 9)(25y^2 - 10y + 1) = 0\) indicates that for the product of two terms to be zero, one of these terms must itself be zero. This is known as the zero-product property.
• Consequently, solving each portion of the equation separately helps in identifying potential roots.

This means we aim to find 'y' values for which either \(4y^2 + 9 = 0\) or \(25y^2 - 10y + 1 = 0\) hold true. For the equation \(4y^2 + 9 = 0\), there are no real roots because the expression under the square root would be negative, which does not yield a real number. Meanwhile, solving the second equation gives us a valid root. This illustrates how knowing and applying the concept of roots can simplify solving complex polynomial equations.

Inspection Method

The inspection method involves "inspecting" or "looking over" the equation to quickly see potential solutions without detailed calculations. This is a useful technique for simple polynomial equations where one can immediately identify what makes the equation zero. Applied here, the inspection method starts with observing that

• for the equation \((4y^2 + 9)(25y^2 - 10y + 1) = 0\), we inspect each factor separately.
• Inspecting \(4y^2 + 9\), we realize quickly it cannot be zero due to the positive nature of squares, which simplifies our task.
• Then, inspection of \(25y^2 - 10y + 1 = 0\) suggests possible factorization or inspection to see if it forms a recognizable pattern, such as a perfect square.

This quick-look approach greatly aids in determining potential solutions faster, especially when combined with a keen eye for recognizing algebraic identities or factored forms in polynomials.

Factoring Expressions

Factoring refers to breaking down an expression into a product of simpler expressions. This is crucial in solving polynomial equations, as it enables us to apply the zero-product property efficiently.
In our example, after deeming \(4y^2 + 9\) non-solvable for real solutions, we focus on the expression \(25y^2 - 10y + 1\). Here, we recognize it as a perfect square trinomial and can rewrite it as:

• \((5y - 1)^2 = 0\)
• This instantly enables us to solve \(5y - 1 = 0\), thus finding the root.

Successfully factoring expressions requires practice in pattern recognition, such as spotting binomials or the difference of squares. It is a powerful tool in any mathematician's toolkit for simplifying and solving equations. This enables a deeper comprehension of how polynomials behave and interact, hence aiding in not just finding solutions, but understanding the underlying algebraic structure.