Problem 35 - Elementary Algebra

Question

Solve by completing the square. $$x 2-23 x-13=0$$

Step-by-Step Solution

Answer

The solutions are approximately \( x \approx 23.55 \) and \( x \approx -0.55 \).

Step 1: Write the equation in standard form

First, ensure the equation is written in the standard quadratic form, which is: \[ ax^2 + bx + c = 0 \]Your equation is already in this form: \[ x^2 - 23x - 13 = 0 \]

Step 2: Move the constant term

Move the constant term (-13) to the other side of the equation: \[ x^2 - 23x = 13 \]

Step 3: Find the number to complete the square

Take the coefficient of the linear term (which is -23), halve it, and then square it: First, find half: \[ \frac{-23}{2} = -11.5 \]Next, square it: \[ (-11.5)^2 = 132.25 \]

Step 4: Add and subtract the square inside the equation

Add and subtract the square you found to the left side to complete the square. Do this carefully to keep the equation balanced: \[ x^2 - 23x + 132.25 - 132.25 = 13 \]

Step 5: Rearrange as a perfect square trinomial

Write the equation as a perfect square and move the subtracted square to the right side:\[ (x - 11.5)^2 = 13 + 132.25 \]

Step 6: Solve for x

Simplify the right side and solve for \( x \):\[ (x - 11.5)^2 = 145.25 \]Take the square root on both sides:\[ x - 11.5 = \pm \sqrt{145.25} \]Finally, solve for \( x \) by adding 11.5:\[ x = 11.5 \pm \sqrt{145.25} \]

Step 7: Simplify the solutions

Evaluate the square root and express the final solutions:\( \sqrt{145.25} \approx 12.05 \) (simplifying the radical can yield more precise results if needed)Thus, \[ x = 11.5 + 12.05 \] or \[ x = 11.5 - 12.05 \]The simplified solutions are:\[ x \approx 23.55 \] and \[ x \approx -0.55 \]

Key Concepts

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable, usually denoted as \(x\).
The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] Where \(a\), \(b\), and \(c\) are constants, and \(a e 0\).
Key features of a quadratic equation include:

• The highest exponent of the variable is 2, indicating it's a quadratic term.
• The graph of a quadratic function is a parabola, which opens upwards if \(a > 0\) and opens downwards if \(a < 0\).
• Solutions to the quadratic equation are also known as the roots, and they may be found using methods like factoring, the quadratic formula, or completing the square.

Understanding the structure of a quadratic equation is crucial for finding its solutions and analyzing the properties of the corresponding parabola.
In the given exercise, the quadratic equation is already in standard form: \(x^2 - 23x - 13 = 0\). This sets the stage for the solution process using different methods including completing the square.

Perfect Square Trinomial

A perfect square trinomial is an expression that can be factored into a square of a binomial.
It takes the form: \[ (ax + b)^2 = a^2x^2 + 2abx + b^2 \] In completing the square, the goal is to rewrite a quadratic expression as a perfect square trinomial.Here’s how it works:

• Start with the quadratic expression \(x^2 + bx\).
• To make it a perfect square trinomial, you need to add \(\left( \frac{b}{2} \right)^2\) to both sides of the equation.
• This process completes the square, converting \(x^2 + bx\) into \(\left( x + \frac{b}{2} \right)^2\).

In the original exercise, by taking \(-23\), halving it to get \(-11.5\), and then squaring to get \(132.25\), we crafted the expression \((x - 11.5)^2\), thus achieving the perfect square trinomial form needed for solving the equation.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \(x\) that satisfy the equation.
One effective method for solving when the equation cannot be factored easily is completing the square.Here are the steps involved in this method:

• Start by ensuring the quadratic equation is in standard form \(ax^2 + bx + c = 0\).
• Move the constant term \(c\) to the right side of the equation.
• Complete the square on the left side by adding \(\left(\frac{b}{2}\right)^2\).
• Express the left side as a square of a binomial.
• Solve for \(x\) by isolating the binomial, taking the square root of both sides, and solving for the variable.

In the exercise, these steps led to the formation of the equation \((x - 11.5)^2 = 145.25\), resulting in solutions by taking the square root and finding \(x = 11.5 \pm \sqrt{145.25}\). This method is powerful as it handles any quadratic equation and provides insights into the nature of the solutions.