Problem 27 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$4 x(x+1)=3$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2}\) and \(x = -\frac{3}{2}\).

Step 1: Expand the Equation

Start by expanding the left side of the equation: \[ 4x(x+1) = 4x^2 + 4x \]Then, set the equation equal to zero by subtracting 3 from both sides: \[ 4x^2 + 4x - 3 = 0 \]

Step 2: Factor the Quadratic Equation

Next, factor the quadratic equation. We need to find two numbers that multiply to \(-12\) (the product of \(4\) and \(-3\)) and add up to \(4\) (the middle coefficient). These numbers are \(6\) and \(-2\). Rewriting the middle term, we have:\[ 4x^2 + 6x - 2x - 3 = 0 \]Group the terms:\[ (4x^2 + 6x) + (-2x - 3) = 0 \]Factor by grouping:\[ 2x(2x + 3) - 1(2x + 3) = 0 \]Factor out the common binomial factor, \(2x + 3\):\[ (2x - 1)(2x + 3) = 0 \]

Step 3: Solve the Factored Equation

Set each factor equal to zero and solve for \(x\):1. \(2x - 1 = 0\) leads to \(2x = 1\) and therefore \(x = \frac{1}{2}\).2. \(2x + 3 = 0\) leads to \(2x = -3\) and therefore \(x = -\frac{3}{2}\).

Key Concepts

Factoring quadratics

Factoring quadratics involves expressing a quadratic equation in a product of simpler polynomials. When we have an equation like \(4x^2 + 4x - 3 = 0\), the goal is to find binomials that multiply together to give the original quadratic. This process can simplify solving the equation significantly.

To begin factoring, it's important to identify two numbers that multiply to the constant term of the quadratic. Here, you multiply the coefficient of the \(x^2\) term (which is \(4\)) by the constant term (which is \(-3\)). This results in \(-12\). Now, you need numbers that not only multiply to \(-12\) but also add up to the coefficient of the \(x\) term, which is \(4\).

• To find these numbers, check the factors of \(-12\): pair them as \(1\) and \(-12\), \(2\) and \(-6\), etc.
• Through trial or calculation, find the numbers \(6\) and \(-2\) that work since \(6 \times -2 = -12\) and \(6 + (-2) = 4\).
• This allows us to rewrite the middle term, effectively preparing for grouping.

By factoring the quadratic, it breaks down into linear components, making it soluble through straightforward algebraic steps.

Algebraic expressions

An algebraic expression is a combination of numbers, variables, and operations like addition, subtraction, multiplication, and division. Our quadratic equation, for instance, begins as \(4x(x+1)=3\). This is a product of an expression \((4x)\) and another expression \((x+1)\).

When working with algebraic expressions:

• Always look to simplify. Simplification helps in identifying the components necessary for solving equations.
• The expression is expanded using algebraic methods, such as distribution, producing \(4x^2 + 4x\) from \(4x(x+1)\).
• It's crucial to keep the equation balanced. Whatever you do to one side, be sure to do to the other side. This keeps the expression equal.

Through these expressions, we define relationships that can be manipulated algebraically to solve for unknown values.

Solving equations

Solving equations is about finding the values of variables that make the equation true. Quadratic equations, like \(4x^2 + 4x - 3 = 0\), can be solved using various methods like factoring, completing the square, or the quadratic formula.

After your equation is factored, as in \((2x - 1)(2x + 3) = 0\), solving becomes a matter of applying the Zero Product Property. This property states that if the product of two factors is zero, at least one of the factors must be zero.

Steps to solve:

• Set each factor equal to zero: \(2x - 1 = 0\) and \(2x + 3 = 0\).
• Solve each equation for \(x\): for \(2x - 1 = 0\), add \(1\) to both sides and then divide by \(2\) to find \(x = \frac{1}{2}\).
• For \(2x + 3 = 0\), subtract \(3\) from both sides and then divide by \(2\) to find \(x = -\frac{3}{2}\).

These solutions represent where the quadratic equals zero on a graph, corresponding to the x-intercepts of the parabola.