Problem 3 - Algebra 2

Question

Solve each equation by using the Square Root Property. \(x^{2}+16 x+64=7\)

Step-by-Step Solution

Answer

The solutions are \(-8 + \sqrt{7}\) and \(-8 - \sqrt{7}\).

Step 1: Move Constant to the Other Side

First, move 64 to the right side of the equation by subtracting 64 from both sides. This gives you\[ x^2 + 16x = 7 - 64 \]which simplifies to\[ x^2 + 16x = -57 \]

Step 2: Complete the Square

To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides. Half of 16 is 8, and 8 squared is 64. Add 64 to both sides:\[ x^2 + 16x + 64 = -57 + 64 \]This becomes:\[ (x + 8)^2 = 7 \]

Step 3: Apply the Square Root Property

Now, take the square root of both sides of the equation using the Square Root Property:\[ \sqrt{(x + 8)^2} = \pm \sqrt{7} \]This gives us:\[ x + 8 = \pm \sqrt{7} \]

Step 4: Solve for x

Finally, isolate \(x\) by subtracting 8 from both sides:\[ x = -8 \pm \sqrt{7} \]So the solutions are:\[ x = -8 + \sqrt{7} \quad \text{and} \quad x = -8 - \sqrt{7} \]

Key Concepts

Completing the Square

Completing the square is a method used in algebra to simplify and solve quadratic equations. It involves creating a perfect square trinomial from a quadratic expression. But why do we want a perfect square? Because it lets us apply the square root property easily.

Here's how you complete the square:

• First, ensure that the coefficient of the quadratic term is 1. If it's not, divide the entire equation by this coefficient.
• Next, look at the linear coefficient, the number in front of the \(x\) term.
• Divide this coefficient by 2, and then square the result. This squared number will be your magic number that you add to both sides of the equation.
• Now, rewrite the quadratic and linear terms along with your magic number as a perfect square trinomial, which can be expressed as \((x + a)^2\).

This transform makes the left side of the equation a perfect square, simplifying the process of solving the equation. It's a very handy technique to handle quadratics, especially when other methods seem cumbersome or complicated.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, which means the highest power of the variable is squared. They take the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving quadratic equations can be approached in several ways.

Key methods include:

• Factoring: This involves expressing the quadratic expression as a product of two binomials. This method works well when the equation can be factored easily.
• Completing the Square: This technique helps to rewrite the quadratic in a perfect square form, which then allows for straightforward solving using the square root.
• Quadratic Formula: Given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), this formula provides solutions directly, making it useful for complex or impossible-to-factor quadratics.

Understanding each method provides flexibility depending on the structure of the quadratic equation you are solving. Whether you choose factoring, completing the square, or the quadratic formula, each approach has its strengths and ideal use cases.

Algebraic Solutions

Algebraic solutions involve manipulating an equation using algebraic properties and operations to isolate the variable of interest. For quadratic equations, employing algebraic solutions typically involves re-arranging terms and leveraging mathematical properties.

To solve quadratics using algebra, we can:

• Move terms across the equal sign to isolate the quadratic, linear, and constant terms appropriately.
• Use operations such as addition, subtraction, multiplication, division, and the square root, to simplify and solve for \(x\).
• Follow systematic steps to maintain balance in the equation, such as completing the square or factoring.
• Cross-check the solutions by plugging back the values of \(x\) into the original equation to ensure they satisfy the equation.

These methods allow us to achieve algebraic solutions, providing both exact numerical values and insights into the equation's behavior. With each method, practice helps solidify understanding and ease problem-solving in various algebraic contexts.