Problem 51 - Algebra 2

Question

Solve each equation by completing the square. \(x^{2}-4.7 x=-2.8\)

Step-by-Step Solution

Answer

The solutions are \(x = 4\) and \(x = 0.7\).

Step 1: Rearrange the Equation

First, we need to move all terms involving the variable to one side. The equation is already in the form of a quadratic equation with the variable terms on one side: x^2 - 4.7x = -2.8.

Step 2: Identify the Term to Complete the Square

Take the coefficient of the linear term (which is -4.7), divide it by 2, then square the result. First, divide it by 2: \[-4.7 \div 2 = -2.35\]Now, square it:\[-2.35^2 = 5.5225\]This is the term we need to add to both sides of the equation to complete the square.

Step 3: Add the Square to Both Sides

Add the term we calculated (5.5225) to both sides of the equation:\[x^2 - 4.7x + 5.5225 = -2.8 + 5.5225\]This simplifies to:\[x^2 - 4.7x + 5.5225 = 2.7225\]

Step 4: Write Left Side as a Perfect Square

Now express the left side of the equation as a perfect square trinomial:\[(x - 2.35)^2 = 2.7225\]We confirm this is correct since \[(x - 2.35)^2 = x^2 - 4.7x + 5.5225\]

Step 5: Solve for x by Taking the Square Root

To solve for \(x\), take the square root of both sides:\[\sqrt{(x - 2.35)^2} = \sqrt{2.7225}\]This yields:\[x - 2.35 = \pm 1.65\]

Step 6: Solve for x

Solve the two resulting equations:1. \(x - 2.35 = 1.65\) \(x = 4\)2. \(x - 2.35 = -1.65\) \(x = 0.7\)The solutions are \(x = 4\) and \(x = 0.7\).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that includes a term with the variable raised to the power of two, or squared. The standard form of a quadratic equation is represented as: \[ax^2 + bx + c = 0\]In this equation, \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The primary feature of a quadratic equation is its parabolic graph, which can open upwards or downwards depending on the sign of the leading coefficient \(a\).
Quadratic equations can have one or two real solutions, or possibly two complex solutions, depending on the discriminant, \(b^2 - 4ac\). In our example, the equation is \(x^2 - 4.7x = -2.8\), which can be rewritten as \(x^2 - 4.7x + 2.8 = 0\) to match the standard form.

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression of the form \((ax + b)^2\) that expands to \(a^2x^2 + 2abx + b^2\). When completing the square in a quadratic equation, our goal is to convert the equation into a perfect square trinomial.
To do this, we take half of the linear coefficient (the coefficient of \(x\)), square it, and add it to both sides of the equation. For example, in our problem, the coefficient of \(x\) is \(-4.7\). We find half of this, \(-2.35\), and square it to get \(5.5225\). Adding \(5.5225\) to both sides transforms the left side into the perfect square trinomial \((x - 2.35)^2\).
Completing the square helps to simplify the equation to easily solve for the variable.

Square Root Method

The square root method is a technique used to solve equations where a square of a binomial is equal to a number. Once we have a perfect square trinomial, like \((x - 2.35)^2 = 2.7225\), we can solve for \(x\) using the square root method.
Here's the process:

• Take the square root on both sides of the equation: \[\sqrt{(x - 2.35)^2} = \sqrt{2.7225}\]
• This simplifies to: \(x - 2.35 = \pm 1.65\)

From here, there will typically be two possible solutions due to the positive and negative square roots, leading us to solve two linear equations.

Algebraic Techniques

Algebraic techniques encompass various methods used to manipulate equations and expressions to find solutions. Completing the square and applying the square root method are both algebraic techniques. These methods allow us to transform a quadratic equation into a more manageable form.
Other important algebraic techniques include:

• Factoring, where a polynomial is expressed as a product of its factors.
• The quadratic formula, a universal method for finding roots of any quadratic equation: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• Simplification, which involves reducing complex expressions to simpler forms.

Each method is suitable for different types of quadratic equations, and selecting the right technique can ease the process of solving the equation.