Solving Quadratic Equations Using the Factoring Method

Introduction

When you encounter a quadratic equation, the factoring method is a powerful tool to solve it. This article explains the steps to solve the quadratic equation (x^2 - 4x - 3 0) using the factoring method. We'll break down the process and provide a shortcut for quick solutions.

Solving the Quadratic Equation Step-by-Step

To solve the quadratic equation (x^2 - 4x - 3 0) using the factoring method, follow these steps:

Step 1: Identify the Quadratic Equation

The equation is in the standard form (ax^2 bx c 0). Here, (a 1), (b -4), and (c -3).

Step 2: Factor the Quadratic

We need to find two numbers that multiply to (ac 1 times -3 -3) and add to (b -4). The numbers that satisfy these conditions are (-1) and (-3).

Step 3: Write the Factored Form

Using these numbers, we can express the quadratic as:

begin{equation*}(x - 1)(x - 3) 0end{equation*}

Step 4: Set Each Factor to Zero

Now, set each factor equal to zero:

begin{equation*}x - 1 0 quad text{or} quad x - 3 0end{equation*}

Step 5: Solve for (x)

Solve for (x) from each equation:

begin{align*}x - 1 0 Rightarrow x 1 x - 3 0 Rightarrow x 3end{align*}

Therefore, the solutions to the equation (x^2 - 4x - 3 0) are (x 1) and (x 3).

Shortcut and Simplified Method

I have a favorite shortcut to solve this problem. Here's an example:

begin{equation*}x^2 - 4x - 3end{equation*}

Convert the equation into a number and factorize the number:

begin{equation*}100 - 40 - 3end{equation*}

The number is (143), and its factors are (11) and (13).

Next, we use the factors to form the quadratic expression:

begin{equation*}x^2 - 1 cdot 25end{equation*}

The number (225 15 times 15), and the factors are negative, which gives us:

begin{equation*}(x - 5)(x 5)end{equation*}

This implies:

begin{align*}x - 5 0 Rightarrow x 5 x 5 0 Rightarrow x -5end{align*}

Identification of Roots and Coefficient

The factors must have (x) multipliers because that is the coefficient of (x^2). Assuming integers, the factors look like (x - ax - b).

Similarly, the product of the constant terms in the factors is (-3).

If the roots are integers, there are only two possibilities: (-1) and (3).

The coefficient of the middle term (-4) is made up of the cross terms or mixed terms of the factors: (-4x -ax - bx -abx). So, (-ab -4) and we must choose (-1) and (3) to match the last term.

Thus, (x^2 - 4x - 3 (x - 1)(x 3)), which will be zero when (x) is a root. Because the factors are different, this implies that one of the factors is zero at a root, so there are two roots (x -1) and (x -3).

Verification Using Grouping Method

We can also verify the solution by rearranging the terms:

begin{equation*}x^2 - x - 3x - 3 0end{equation*}

Group the terms as follows:

begin{equation*}x(x - 1) - 3(x - 1) 0end{equation*}

Factoring out the common term ((x - 1)), we get: