Question

MULTIPLE CHOICE Solve \(x^{2}+8 x-2=0\) $$\begin{array}{ccccc}\mathbf{A})-4 \pm 3 \sqrt{2} & \mathbf{B} & 4 \pm 3 \sqrt{2} & \mathbf{C} & -4 \pm 2 \sqrt{2} & \mathbf{D} & \mathbf{4} \pm \sqrt{\mathbf{1 6}} \end{array}$$

Short answer

The correct answer is \(x = -4 \pm 3\sqrt{2}\), which corresponds to option A.

Step-by-step solution

Identify values of a, b and c

From the quadratic equation \(x^{2}+8 x-2=0\), identify the values of a, b and c. Therefore, a=1, b=8, c=-2.

Apply the quadratic formula

Substitute the values into the quadratic formula, which is given by \(-b \pm \sqrt{b^{2}-4ac}/2a\). Doing this gives us \(x = -8 \pm \sqrt{(8)^{2}-4(1)(-2)}/2(1)\). Simplify this to get \(x = -8 \pm \sqrt{64+8}/2\).

Simplify the result

Simplify under the square root to get \(x = -8 \pm \sqrt{72}/2\). Simplify further to get \(x = -4 \pm \sqrt{18}\), or \(x = -4 \pm 3\sqrt{2}\).

Match the result with the provided options

The obtained solution, \(x = -4 \pm 3\sqrt{2}\), matches with option A.

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra that allows us to find the roots of any quadratic equation, which is any equation in the form of \(ax^2 + bx + c = 0\). It is derived from the process of completing the square and is given by the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula offers a systematic way to calculate the solutions to a quadratic equation by plugging the numerical coefficients of the equation into the formula. The symbol \(\pm\) indicates that there are usually two solutions to a quadratic equation: one for the addition and one for the subtraction of the square root term. In the exercise \(x^2 + 8x - 2 = 0\), the application of the quadratic formula simplified the problem by providing a direct route to the answer, \(-4 \pm 3\sqrt{2}\), corresponding to option A.

Factoring Quadratics

Factoring quadratics is another method to solve quadratic equations, and it involves expressing the quadratic equation as the product of two binomial expressions. This method is used when a quadratic can be written in the form \((px + q)(rx + s) = 0\), where \(p, q, r,\) and \(s\) are numbers that, when multiplied, give the original quadratic coefficients.

The solution to the equation exists when one or both of these binomials equal zero, based on the Zero Product Property which states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\) or both. In some cases, like with the exercise's equation \(x^2 + 8x - 2 = 0\), factoring may not be straightforward, and thus the quadratic formula presents a more feasible solution.

Square Roots in Algebra

Square roots in algebra represent a fundamental operation which involves finding a number that, when multiplied by itself, will yield the original number under the square root sign. In the context of solving quadratic equations, square roots can help simplify the process once the quadratic formula has been applied.

For instance, in our original exercise, after applying the quadratic formula, we encountered the square root of 72. This can be further simplified by identifying factors of 72 that are perfect squares, such as 36. This process helps to simplify the radical \(\sqrt{72}\) into \(\sqrt{36 \times 2}\), which then simplifies to \(6\sqrt{2}\). Simplification of square roots is essential to arriving at the most straightforward form of the solutions.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. Understanding how to manipulate these expressions is key to solving algebraic equations, including quadratics. An expression such as \(x^2 + 8x - 2\) in our exercise is an algebraic expression where \(x\) is the variable, and the numbers 8 and -2 are coefficients and constants, respectively.

When dealing with algebraic expressions, it is important to recognize the order of operations and how to simplify expressions correctly. For example, simplifying the result of the quadratic formula involves combining like terms and reducing fractions where possible, leading to the solution of the equation.