Problem 109 - Introductory Algebra for College Students

Question

Intercepts for the graph in \(a[-10,10,1]\) by \([-13,10,1]\) viewing rectangle to solve the quadratic equation. Check by substitution. Use the graph of \(y=(x-2)(x+3)-6\) to solve $$(x-2)(x+3)-6=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \((x-2)(x+3)-6=0\) are \(x = 2\) and \(x = -3\), as confirmed by substitution.

Step 1: Expand and simplify

The expression \((x-2)(x+3)-6=0\) can be expanded and simplified to \(x^2 + x - 12 = 0\) using the distributive property.

Step 2: Solving the Quadratic Equation

Once the equation is in the form \(ax^2 + bx + c = 0\), apply the quadratic formula \(x = [-b \pm sqrt(b^2 - 4ac)] / 2a\) where a, b, and c are coefficients from the equation (\(a = 1, b = 1, c = -12\)). Thus the roots of the equation can be found. For this problem, the roots are \(x = 2\) and \(x = -3\).

Step 3: Verification

To confirm the roots, plug values of x into the original equation. If the result is zero, the roots have been correctly solved. Substituting \(x = 2\) in \((x-2)(x+3)-6=0\), the result is zero. Similarly, substituting \(x = -3\) produces zero too. Both roots are verified.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation, which is a polynomial equation of the form \(ax^2 + bx + c = 0\). The formula is given by:

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

In this equation:

• \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation.
• \(b^2 - 4ac\) is called the discriminant, which determines the nature of the roots.

If you plug \(a = 1\), \(b = 1\), and \(c = -12\) into the formula, you can solve for the roots of the equation \(x^2 + x - 12 = 0\). The roots are the values of \(x\) where the quadratic equation equals zero, and they are \(x = 2\) and \(x = -3\) in this case. The formula simplifies finding solutions to quadratic equations, especially when factoring is difficult.

Distributive Property

The distributive property is one of the most fundamental properties in algebra. It states that for all numbers \(a\), \(b\), and \(c\), the following holds:

• \(a(b + c) = ab + ac\)

This property is used to expand expressions by multiplying each term inside a parenthesis by the term outside. Let's take a look at how it's applied to the expression \((x-2)(x+3)\) from our problem:

• First, distribute \(x\): \(x \cdot (x+3) = x^2 + 3x\)
• Next, distribute \(-2\): \(-2 \cdot (x+3) = -2x - 6\)

Add the results together to get: \(x^2 + 3x - 2x - 6\), which simplifies to \(x^2 + x - 6\).The distributive property helps simplify polynomial expressions and is key in transforming the initial expression into a standard quadratic form. In this exercise, it allowed the transition from \((x-2)(x+3)-6\) to the simplified equation \(x^2 + x - 12 = 0\).

Verifying Roots

Verifying the roots of a quadratic equation is an important step to ensure the accuracy of the solutions. This involves substituting the obtained roots back into the original equation to see if they satisfy the equation, giving a result of zero. Here's how it's done:
Let's say the original equation is \((x-2)(x+3)-6=0\) and the roots are \(x = 2\) and \(x = -3\). Verification requires two steps:

• Substitute \(x = 2\): \((2 - 2)(2 + 3) - 6 = 0\Rightarrow 0 \cdot 5 - 6 = 0\). This confirms that \(x = 2\) is a root.
• Substitute \(x = -3\): \((-3 - 2)(-3 + 3) - 6 = 0\Rightarrow -5 \cdot 0 - 6 = 0\). This confirms that \(x = -3\) is also a root.

If both substitutions produce zero, the roots are correctly verified. Verifying roots is a crucial final check that helps confirm the solutions are correct and ensures no mistakes were made during calculations. This step prevents errors from going unnoticed in mathematical problem-solving.