Problem 41 - Intermediate Algebra

Question

For Exercises 37 through \(42,\) first divide the equation through by the coefficient of \(\left.x^{2} \text { (or } y^{2}\right)\) $$4(x+1)^{2}+4(y-3)^{2}=12$$

Step-by-Step Solution

Answer

\((x+1)^{2} + (y-3)^{2} = 3\)

Step 1: Identify the Coefficient of the Square

The given equation is \(4(x+1)^{2}+4(y-3)^{2}=12\). Observe that the coefficients of both \((x+1)^{2}\) and \((y-3)^{2}\) is 4.

Step 2: Divide the Entire Equation by the Coefficient

Divide each term of the equation \(4(x+1)^{2}+4(y-3)^{2}=12\) by 4, which is the common coefficient of \((x+1)^{2}\) and \((y-3)^{2}\). This results in: \((x+1)^{2} + (y-3)^{2} = 3\).

Step 3: Simplify the Resulting Equation

After division, the equation simplifies to \((x+1)^{2} + (y-3)^{2} = 3\). There are no further simplifications to perform as the equation is already simplified.

Key Concepts

Dividing Polynomials

When we talk about dividing polynomials, we generally mean simplifying polynomial expressions by division. In this context, it involves dividing every term in the equation by a common factor, often a coefficient of the leading term. This process is similar to factoring in arithmetic, where you divide numbers by their greatest common divisor (GCD).

Here's how it works:

• Look at the equation and identify a common coefficient shared by all terms involving polynomials.
• Divide every term in the equation by this coefficient to simplify it, essentially redistributing the values evenly.

For example, in our original exercise, we started with an equation where each term involving squares had a coefficient of 4. By dividing each term by 4, we reduced the equation to a simpler form. This simplification allows for further manipulation or easier analysis, such as graphing a circle in this case.

Quadratic Equations

Quadratic equations are defined as polynomial equations of degree two, typically in the form of \(ax^2 + bx + c = 0\). However, they can also appear in different configurations, such as when expressed in vertex form like \( (x+h)^2 = k \).

The most common techniques for solving quadratic equations involve:

• Factoring: Writing the equation in a factored form to find the roots.
• Using the Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Completing the Square: Adjusting the equation into a perfect square trinomial.
• Graphing: Finding solutions by interpreting the parabola's x-intercepts.

In our context, by simplifying the equations involving squares to a form without coefficients, we could more easily compare it to a standard geometric shape. This shows how even quadratic terms can simplify seemingly complex equations.

Algebraic Simplification

Algebraic simplification involves reducing expressions into their simplest form without changing their value. This process makes equations easier to solve or interpret and can be incredibly useful in a wide range of algebraic problems.

In algebraic simplification:

• Identify and factor out common terms or coefficients.
• Perform operations such as addition, subtraction, multiplication, or division evenly across the equation.
• Combine like terms and eliminate any redundant expressions.

Returning to our problem, by dividing each term by the coefficient of 4, the complex equation was simplified into a more manageable format. Simplified equations are not only easier to work with mathematically but also if you're visually interpreting them, such as recognizing it as the equation of a circle, as we saw in the exercise.