Problem 46 - Brief Applied Calculus

Question

Solve each equation by factoring or the Quadratic Formula, as appropriate. $$ 3 x^{2}-6 x+9=6 $$

Step-by-Step Solution

Answer

The solution is \(x = 1\).

Step 1: Move all terms to one side

First, let's start by setting the equation in standard quadratic form. We need to move all terms to one side of the equation:\[3x^2 - 6x + 9 = 6\]Subtract 6 from both sides:\[3x^2 - 6x + 9 - 6 = 0\]This simplifies to:\[3x^2 - 6x + 3 = 0\]

Step 2: Identify and simplify the quadratic equation

Look at the quadratic equation we have:\[3x^2 - 6x + 3 = 0\]Notice that all terms can be divided by 3 to simplify the equation:\[x^2 - 2x + 1 = 0\]

Step 3: Factor the quadratic equation

Now we'll factor the quadratic equation. We look at:\[x^2 - 2x + 1 = 0\]The expression is a perfect square trinomial. It can be factored as:\[(x - 1)^2 = 0\]

Step 4: Solve for the variable

We solve the factored equation:\[(x - 1)^2 = 0\]To solve, we take the square root of both sides to get:\[x - 1 = 0\]Then, add 1 to both sides:\[x = 1\]

Key Concepts

Factoring Quadratics

Understanding how to factor quadratics is crucial for solving certain quadratic equations. In such cases, we write the quadratic as a product of two binomials. This can make the process of finding solutions much simpler.
For a quadratic in standard form, such as \[ax^2 + bx + c = 0,\]our aim is to express it as \[(x - p)(x - q) = 0.\]This means that if we can find numbers \(p\) and \(q\) such that \(pq = c\) and \(p + q = b,\) then the quadratic can be factored.

• The equation \(x^2 - 2x + 1 = 0\) is an example of a factorable quadratic.
• By recognizing it as a perfect square trinomial, it mellows into \((x - 1)^2 = 0.\)

Factoring quadratics this way, especially recognizing patterns like perfect square trinomials, can be a powerful technique to efficiently solve quadratic equations.

Quadratic Formula

When factoring is complex or not feasible, the quadratic formula is our reliable friend. It provides a surefire method to find the roots of any quadratic equation. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Using this formula requires the coefficients \(a\), \(b\), and \(c\) from the quadratic equation in the standard form \(ax^2 + bx + c = 0\).

• This formula helps when the quadratic doesn't factor neatly.
• The term under the square root, \(b^2 - 4ac\), is called the discriminant.
• The discriminant tells us about the nature of the roots: two distinct real roots, one real root, or no real roots.

The quadratic formula can solve any quadratic equation, making it a fundamental part of algebra.

Perfect Square Trinomials

A perfect square trinomial is a quadratic that can be represented as the square of a binomial. It has a particular structure, which allows for recognizing and factoring them efficiently. The general forms of perfect square trinomials are:

• \((a + b)^2 = a^2 + 2ab + b^2,\) and
• \((a - b)^2 = a^2 - 2ab + b^2.\)

Identifying perfect square trinomials allows for immediate factoring, simplifying the solving process. In our equation \(x^2 - 2x + 1 = 0,\) we spotted that it matches the pattern:
\[(x - 1)^2 = x^2 - 2x + 1.\]Recognizing such a pattern:

• Saves time when solving equations.
• Prepares students to solve more complex problems by breaking them down into simple parts.

This understanding helps students work through and solve equations efficiently, just like in our example.