Problem 117 - Intermediate Algebra

Question

Solve each equation. $$a(a-6)=-9$$

Step-by-Step Solution

Answer

The solution is \(a = 3\).

Step 1: Expand the equation

The given equation is \(a(a-6) = -9\). Start by expanding the left-hand side of the equation. Distribute \(a\) over \(a-6\) to get \(a^2 - 6a = -9\).

Step 2: Move terms to one side

Rearrange the equation by moving all terms to one side to form a quadratic equation, \(a^2 - 6a + 9 = 0\).

Step 3: Identify the quadratic form

The equation \(a^2 - 6a + 9 = 0\) is now a standard quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -6\), and \(c = 9\).

Step 4: Factor the quadratic

Factor the quadratic equation. Notice that \(a^2 - 6a + 9\) can be factored as \((a-3)(a-3) = 0\) or \((a-3)^2 = 0\).

Step 5: Solve for \(a\)

Since \((a-3)^2 = 0\), set \(a-3 = 0\). Solve for \(a\) to find \(a = 3\).

Key Concepts

Factoring Quadratics

Factoring quadratics is a crucial step in solving some quadratic equations. A quadratic equation is often written in the form of \(ax^2 + bx + c = 0\).
When you "factor" a quadratic, you try to rewrite it as a product of its binomials.

• This is especially helpful because, if the product equals zero, then one or both of the factors must also equal zero.
• The relationship is key because it allows us to find the roots or zeros of the quadratic equation.

In the example given, \(a^2 - 6a + 9 = (a - 3)(a - 3)\).
By factoring, we see that both terms \(a - 3\) are identical, meaning this equation can also be written as \((a - 3)^2 = 0\).
This makes it very clear what the solution should be.

Solving Quadratic Equations

Solving quadratic equations often involves setting the factored form or the expanded form equal to zero.
The goal is to find the variable value(s) that satisfy this. In the example equation, once factored, it is clear that the solution lies where the binomials are zero.

• Once in the form \((a - 3)(a - 3) = 0\), solve each factor by setting them equal to zero.
• Since both are \(a - 3 = 0\), simply solve for \(a\).

So, you have \(a - 3 = 0\), leading to \(a = 3\).
This means 3 is the solution to the quadratic equation, specifically a double root or repeated solution, as both factors are the same.

Quadratic Formula

Though not directly used in the given problem, the quadratic formula is another powerful tool for solving quadratic equations. It is particularly handy when factoring is not straightforward or possible.
The quadratic formula is given by:\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the roots of any quadratic equation in the form \(ax^2 + bx + c = 0\).

• The discriminant \(b^2 - 4ac\) under the square root determines the nature of the roots.
• If it's positive, you get two distinct real roots; if zero, one real double root; and if negative, two complex roots.

This is beneficial because it means you can solve any quadratic, even if it cannot be factored easily.

Polynomials

A polynomial is an algebraic expression consisting of variables raised to whole number exponents, and multiplied or summed with coefficients. In the context of this exercise, a quadratic is a specific type of polynomial with a degree of 2.
When solving polynomial equations, especially quadratics, understanding their properties is vital. Polynomials are essential in algebra and other areas because they:

• Help describe various relationships and functions mathematically.
• Can often be graphed to show their behavior visually, revealing roots, intercepts, and other important information.

The given equation \(a(a-6) = -9\) transforms into the polynomial \(a^2 - 6a + 9\), demonstrating how manipulation of equations converts them into standard polynomial forms, making them easier to solve through familiar techniques.