Problem 46 - Algebra A Combined Function

Question

Solve each equation by completing the square. $$ x^{2}-10 x+2=0 $$

Step-by-Step Solution

Answer

x = 5 ± √23.

Step 1: Move the constant term

Begin by isolating the quadratic and linear terms. Move the constant term to the other side of the equation:\[ x^2 - 10x = -2 \]

Step 2: Prepare to complete the square

Take the coefficient of the linear term (which is -10 in this case), divide it by 2, and square the result. - The term is \(-10\).- Dividing by 2 gives \(-5\).- Squaring gives \((-5)^2 = 25\).

Step 3: Add and subtract the square term

Add this square term (25) to both sides of the equation to maintain balance:\[ x^2 - 10x + 25 = 23 \]Now the left side is a perfect square trinomial.

Step 4: Factor the perfect square trinomial

The left-hand side of the equation is now a perfect square trinomial, which can be factored as:\[ (x - 5)^2 = 23 \]

Step 5: Solve for x by taking the square root

Take the square root of both sides to solve for \(x\):\[ x - 5 = \pm \sqrt{23} \]This means \(x = 5 \pm \sqrt{23}\).

Step 6: Write the final solution

The solution to the equation is the set of values \(x = 5 + \sqrt{23}\) and \(x = 5 - \sqrt{23}\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The term \( ax^2 \) is called the quadratic term, \( bx \) is the linear term, and \( c \) is the constant term. Quadratic equations model many real-world scenarios, such as the trajectory of a thrown ball or the area of a rectangle given its dimensions.
To solve quadratic equations, there are several methods available:

• Factoring: Expressing the quadratic as a product of binomials.
• Completing the square: Modifying the equation so that the expression becomes a perfect square trinomial.
• Quadratic formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Graphically: Plotting the quadratic function and identifying the x-intercepts.

Each method has its advantages, but completing the square can be particularly insightful for understanding the behavior of the quadratic function.

Perfect Square Trinomial

A perfect square trinomial is a specific type of polynomial that can be written in the form \( (x + a)^2 = x^2 + 2ax + a^2 \).
In the process of completing the square, our goal is to transform a quadratic equation into a perfect square trinomial.
This makes it easier to factor and solve. For example, from the original exercise:

• Start with the quadratic part: \( x^2 - 10x \).
• To complete the square, take half of the linear coefficient (\(-10\)), which is \(-5\), and square it to get \(25\).
• Add \(25\) to both sides to maintain equality.

This process converts \( x^2 - 10x + 25 \) into the perfect square \( (x - 5)^2 \), making it straightforward to solve by taking square roots.

Factoring

Factoring involves rewriting a polynomial as a product of simpler polynomials, or factors. It is a way of breaking down the equation into manageable pieces, which can then be easily solved, especially if the factors are simple binomials.
In solving the original exercise using completing the square, the factoring step transforms the quadratic expression \( x^2 - 10x + 25 \) into \( (x - 5)^2 \).
Factors of a perfect square trinomial can be directly determined:

• Identify the perfect square trinomial: \( x^2 - 10x + 25 \).
• Recognize and express it as \( (x - 5)(x - 5) \) or \( (x - 5)^2 \).

This simplifies solving quadratic equations by reducing them to an equation that can be easily undone using the square root method.

Square Root Method

The square root method is a straightforward approach used to solve equations where the variable is contained within a squared expression. This method is most effective after transforming a quadratic equation into a perfect square form.
Here's how it works in the context of our original problem:

• After completing the square, the equation \( (x - 5)^2 = 23 \) is obtained.
• The square root method involves taking the square root of both sides: \( x - 5 = \pm \sqrt{23} \).
• The solution is obtained by isolating \( x \): \( x = 5 \pm \sqrt{23} \).

This provides the two solutions for \( x \).
The square root method is powerful because it simplifies complex quadratic equations into more manageable linear equations. It vividly demonstrates the concept of equality by maintaining a balance as operations are performed on both sides.