Problem 47 - Algebra 1

Question

Solve the equation by completing the square. $$4 x^{2}+6 x-6=2$$

Step-by-Step Solution

Answer

So, the solutions for \(x\) are approximately \(0.85\) and \(-2.35\).

Step 1: Re-arrange the equation

Start by moving the \(6\) from the left side, which doesn't contain \(x\), to the right side to simplify the expression: \(4x^2 + 6x = 2 + 6\), which simplifies to \(4x^2 + 6x = 8\).

Step 2: Completing the square

To make completing the square easier, divide the entire equation by \(4\) to get \(x^2 + 1.5x = 2\). Then, find the number that completes the square. You do this by halving the coefficient of \(x\), squaring it and adding it to both sides. The coefficient of \(x\) is \(1.5\), after halving and squaring it \((0.75)^2\) we get \(0.5625\). So, the equation becomes: \(x^2 + 1.5x + 0.5625 = 2 + 0.5625\).

Step 3: Simplify and translate into square of binomial

The left side of the equation can now be written as a binomial square and right side simplifies to \(2.5625\). So, the equation becomes: \((x+0.75)^2 = 2.5625\).

Step 4: Solve for x

To isolate \(x\), start by taking the square root of both sides: \(x+0.75 = \pm \sqrt{2.5625}\). This yields two solutions when simplified: \(x \approx -0.75 + 1.6\) and \(x \approx -0.75 - 1.6\).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are central to algebra and appear frequently in various fields such as physics, engineering, and economics. In our example, the equation is \(4x^2 + 6x - 6 = 2\). Notice, it initially doesn't fit the standard form due to the problem being set as an equation rather than equal to zero. Our first step was to rearrange it, clarifying its quadratic nature. Understanding the structure of quadratic equations allows us to choose the correct solving strategy, like completing the square, factoring, or using the quadratic formula.

Solving Equations

Solving equations involves finding the values of the variables that satisfy the given conditions. Here, we aim to find the value of \(x\) for the equation \(4x^2 + 6x = 8\). Solving often requires transforming the original equation using various methods to make it easier to work with. One of the central strategies for solving quadratic equations is re-arranging terms, as seen in the first step where we transformed the equation to isolate squared terms. This helps in simplifying the equation to a form where techniques like completing the square can be applied efficiently.

Algebraic Manipulation

Algebraic manipulation is crucial for simplifying complex equations and expressions. It involves changing the form of an expression using algebraic rules and operations while keeping its equivalence. Examples include re-arranging terms, combining like terms, or dividing by a number to simplify an equation. For instance, dividing the equation \(4x^2 + 6x = 8\) by 4 simplifies it to \(x^2 + 1.5x = 2\). This step is vital as it prepares the equation for completing the square by making the coefficient of \(x^2\) equal to 1, which is a key condition for this method. Carefully executing algebraic manipulations ensures that the original relationship of the equation is maintained, paving the way for easier solutions.

Factoring

Factoring is one of the primary methods of solving quadratic equations, where we express the quadratic as a product of its binomials. However, not all quadratic equations are easily factorable, and hence methods like completing the square are used. The idea of completing the square is related to factoring because you transform the equation so that one side is a perfect square trinomial, which translates to a factored form such as \((x + p)^2\). This method provides us with a useful alternative, especially when initial attempts at simple factoring do not work. In solving \(x^2 + 1.5x + 0.5625 = 2.5625\), it allowed us to transform the equation into \((x + 0.75)^2 = 2.5625\), making it straightforward to apply the square root, bypassing direct factoring.