Problem 10 - Algebra and Trigonometry

Question

Solve the equation by factoring. $$ 4 x^{2}-4 x-15=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = -\frac{3}{2}\) and \(x = \frac{5}{2}\).

Step 1: Identify and Set Up the Quadratic Equation

The given equation is \(4x^2 - 4x - 15 = 0\). Our goal is to factor this quadratic equation to find the values of \(x\) that satisfy it.

Step 2: Setup for Factoring by Splitting the Middle Term

To factor the quadratic expression, we need two numbers whose product is the product of the coefficient of \(x^2\) term (4) and the constant term (-15), which is \(4 \times -15 = -60\), and whose sum is the coefficient of \(x\) term, which is -4.

Step 3: Determine the Pair of Numbers

We need numbers that multiply to -60 and add up to -4. The numbers -10 and 6 meet these criteria: \(-10 \times 6 = -60\) and \(-10 + 6 = -4\).

Step 4: Rewrite the Middle Term Using the Pair of Numbers

Rewrite the quadratic expression as \(4x^2 - 10x + 6x - 15 = 0\). This uses the numbers -10 and 6 found in Step 3.

Step 5: Factor by Grouping

Group the terms: \((4x^2 - 10x) + (6x - 15) = 0\). Factor each group: \(2x(2x - 5) + 3(2x - 5) = 0\).

Step 6: Factor the Common Binomial

The expression \(2x(2x - 5) + 3(2x - 5)\) can be factored further as \((2x + 3)(2x - 5) = 0\) using the common binomial \((2x - 5)\).

Step 7: Solve for Roots

To find the values of \(x\), set each factor equal to zero and solve: \(2x + 3 = 0\) yields \(x = -\frac{3}{2}\) and \(2x - 5 = 0\) yields \(x = \frac{5}{2}\).

Key Concepts

Quadratic Equation

Quadratic equations are fascinating and powerful mathematical tools that appear in various mathematical and real-world contexts. A quadratic equation can be represented as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. The leading coefficient \(a\) should not be zero, as this would turn the equation into a linear one. Its characteristic U-shaped graph is known as a parabola, and the points where this graph cuts the \(x\)-axis are called the roots of the equation. To solve a quadratic equation, you can use different methods:

• Factoring
• Completing the square
• Using the quadratic formula
• Graphing

Each method has its own advantages, but factoring is often preferred when the equation can be easily rewritten into a product of simpler expressions involving \(x\). This approach helps to quickly identify the equation's roots. In the exercise, we start by rearranging the equation into a format that makes it easier to factor, using a strategy called 'factoring by grouping', which is a clever approach to identify those roots.

Roots of an Equation

The roots of an equation are simply the solutions to that equation. For a quadratic equation, these roots are where the parabola intersects the \(x\)-axis. This intersection is critical because it tells us the value of \(x\) that satisfies the equation, essentially the solution of the equation.There are typically two roots in a quadratic equation, but sometimes both can be equal, or the equation might not have real roots. In cases like our example, multiple real roots exist: \(x = \frac{5}{2}\) and \(x = -\frac{3}{2}\).To find these roots:

• Set the factored equation equal to zero.
• Each factor then gives rise to its own simple linear equation.
• Solve these equations to determine the precise value of \(x\).

In the given exercise, the factorization leads to two simple equations: \(2x + 3 = 0\) and \(2x - 5 = 0\) that can be solved to find the values of \(x\), which are the roots of the original quadratic equation.

Factoring by Grouping

Factoring by grouping is a strategic step that simplifies complex polynomial expressions into manageable parts. It is particularly useful in this quadratic equation context when direct factoring is not apparent.Here's how to effectively use this method:

• First, rewrite the quadratic expression by breaking the middle term into two terms, using numbers that multiply to the product of the first and last coefficients, and add up to the middle coefficient.
• Then, reorganize the expression to facilitate easy grouping.
• Next, group the terms into pairs that can be factored more easily.
• Factor out the greatest common factor in each group.
• Identify and extract the common binomial factor from the grouped expression.

In the original problem, after finding the appropriate pair of numbers (-10 and 6), we rewrite \(-4x\) as \(-10x + 6x\), achieving two groups: \((4x^2 - 10x)\) and \((6x - 15)\). Factoring each group separately yields common terms, which enables further simplification to reach the ultimate factorized form \((2x + 3)(2x - 5)\). This step is vital as it unlocks the path to discovering the equation's roots.