Problem 55 - Algebra and Trigonometry

Question

The given equation involves a power of the variable. Find all real solutions of the equation. \(8 x^{2}-64=0\)

Step-by-Step Solution

Answer

The real solutions are \(x = 2\sqrt{2}\) and \(x = -2\sqrt{2}\).

Step 1: Isolate the Quadratic Term

The given equation is \(8x^2 - 64 = 0\). First, add 64 to both sides of the equation to isolate the quadratic term \(8x^2\). This gives us:\[8x^2 = 64\]

Step 2: Simplify the Equation

Next, divide every term in the equation by 8 to simplify the expression. This yields:\[x^2 = 8\]

Step 3: Solve for x by Taking the Square Root

To solve for \(x\), take the square root of both sides of the equation. Remember that taking the square root of a squared term introduces both a positive and negative solution. Thus, we have:\[x = \pm\sqrt{8}\]

Step 4: Simplify the Square Root

Simplify the square root \(\sqrt{8}\). We know that \(8 = 4 \times 2\), and \(\sqrt{4} = 2\), so we can write:\[\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\]

Step 5: Write the Final Solution

Substituting back the simplified square root from Step 4, we have the solutions for \(x\):\[x = 2\sqrt{2} \text{ or } x = -2\sqrt{2}\]

Key Concepts

Solving Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest degree term is squared, or raised to the power of two. Each quadratic equation can typically be expressed in the standard form: \[ ax^2 + bx + c = 0 \] To solve these equations, we aim to find the values of \(x\) that make the equation true. These values are known as the roots or solutions of the quadratic equation. In our case, the equation is:\[ 8x^2 - 64 = 0 \]To solve such equations efficiently:

• First, isolate the quadratic term by moving all non-variable terms to one side, aiming to get a term like \(ax^2\).
• Then, simplify the equation by dividing through by any coefficient of the squared term, if necessary, to make it easier to solve, reducing it to a form such as \(x^2 = ext{number}\).
• Finally, solve for \(x\) by taking the square root of both sides of your simplified equation.

For our example equation, we added 64 to both sides, and then divided by 8, resulting in \(x^2 = 8\). From this, we can take the square root.

Square Roots

The square root is a mathematical operation that, when applied to a number, gives the value that, when multiplied by itself, gives the original number. Mathematically represented as \(\sqrt{}\), the square root of \(n\) is the number that when squared equals \(n\).For example:

• The square root of 9 is 3 because 3 multiplied by 3 equals 9, i.e., \(\sqrt{9} = 3\).
• When finding the square roots during solving quadratic equations, remember that both positive and negative roots exist. Hence, \(x^2 = 9\) results in \(x = 3\) or \(x = -3\).

In the original exercise, solving \(x^2 = 8\) involves taking the square roots on both sides, leading to \(x = \pm \sqrt{8}\). It is key to recall that both the positive and negative square roots are valid solutions.

Simplifying Radicals

Simplifying radicals, like \(\sqrt{8}\) in the given problem, is all about expressing the root in its simplest form. Radicals can often be simplified by factoring the number under the square root into its prime constituents.For instance, with \(\sqrt{8}\), we know that 8 can be expressed as the product of 4 and 2:

• First, factor the number: 8 = 4 \(\times\) 2.
• Next, recognize that \(\sqrt{4}\) is 2 because 2 squared equals 4.
• Thus, \(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\).

By simplifying the radicals, the equation's solutions become clearer and appear in their simplest form. In our case, the solutions for \(x\) after simplifying are \(x = 2\sqrt{2}\) or \(x = -2\sqrt{2}\). Simplifying helps in communicating solutions clearly and makes further computations easier if needed.