Problem 14 - Algebra 1: Concepts and Skills

Question

Solve the equation or write no real solution. $$ 2 x^{2}-8=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2\) and \(x = -2\).

Step 1: Rewrite the Equation

The first step is rewriting the equation to make it easier to understand or simplify it. In this case, the equation is already simplified. So, the rewritten form is \(2 x^{2} - 8 = 0\).

Step 2: Factorize the Equation

Take out the common factors from the equation. In this case, 2 is the common factor. After factoring, expression should look like \(2(x^{2} - 4) = 0\). Then, the inside brackets can be written as a difference of two squares (in the form \(a^{2}-b^{2}\)) as \((x - 2)(x + 2)\). So, after factoring, the equation becomes \(2(x - 2)(x + 2) = 0\).

Step 3: Solve for x

The next step is to set each factor equal to zero and solve for \(x\). Setting \(a(x - 2)(x + 2) = 0\), we get \(x - 2 = 0\) and \(x + 2 = 0\). Solving these gives \(x = 2\) and \(x = -2\) respectively.

Key Concepts

Factoring Quadratics

Mastering the art of factoring quadratics is vital for solving second-degree polynomial equations, which are of the form ax^2 + bx + c = 0. A quadratic equation can often be factored into the product of two binomials, if you can find the magic pair of numbers that work.

For example, in the equation 2x^2 - 8 = 0, you initially look for any common factors of all terms. Here, the number 2 is a common factor, and once factored out, the quadratic x^2 - 4 is what remains. This unique expression is then recognized as the 'difference of two squares' and is further broken down into (x - 2)(x + 2), showcasing the fact that the product of these binomials equates back to the original quadratic expression. Finally, apply the zero product property: if the product of two factors is zero, at least one of the factors must be zero. This paves the way for finding the roots of the quadratic equation.

Difference of Two Squares

The 'difference of two squares' is a special pattern in algebra where a squared term is subtracted from another squared term, expressed as a^2 - b^2. An expression fitting this pattern factors neatly into (a - b)(a + b). It's essential to recognize this pattern as it simplifies solving quadratics significantly.

In the original problem, the equation 2(x^2 - 4) represents a perfect setup for this technique since x^2 and 4 are both squares. Applying the formula for the difference of two squares, we can rewrite x^2 - 4 as (x - 2)(x + 2), paving the path to easily find the solutions to the equation.

Understanding this concept allows students to efficiently factor expressions of this form without lengthy trial and error, enhancing their problem-solving speed and efficiency.

Quadratic Formula

When factoring is challenging or quadratics are complex, the quadratic formula is a reliable and universal method to find the solutions. This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is derived from the coefficients of a quadratic equation ax^2 + bx + c = 0. The symbol \(\pm\) indicates there are usually two solutions.

To apply this powerful tool, identify the values of a, b, and c from your quadratic equation. Substitute these into your formula and calculate. Even if an equation is factorable, the quadratic formula will still work, providing a solid verification for your factored solutions.

The genius of this formula lies in its versatility. No matter the quadratic, as long as it's in standard form and has real solutions, the quadratic formula delivers. Understanding and applying this formula is a valuable skill in a student's mathematical toolkit.