Problem 75 - Elementary Algebra

Question

Problem: Use the quadratic formula to solve \(x^{2}-30=0\). Incorrect Answer: \(x=\frac{-(-30) \pm \sqrt{(-30)^{2}-4(1)(0)}}{2(1)}\) \(x=\frac{30 \pm \sqrt{900-0}}{2}\) \(x=\frac{30 \pm \sqrt{900}}{2}\) \(x=\frac{30 \pm 30}{2}\) \(x=\frac{30+30}{2}\) or \(x=\frac{30-30}{2}\) \(x=\frac{60}{2} \quad\) or \(\quad x=\frac{0}{2}\) \(x=30 \quad\) or \(\quad x=0\)

Step-by-Step Solution

Answer

The solutions to the equation are \( x = \sqrt{30} \) and \( x = -\sqrt{30} \).

Step 1: Identify coefficients

Rewrite the quadratic equation in the standard form: \[x^{2} - 30 = 0 \] Here, \(a=1\), \(b=0\), and \(c=-30\).

Step 2: Write down the quadratic formula

The quadratic formula to solve for \(x\) is: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Step 3: Substitute coefficients into the formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula: \[x = \frac{-(0) \pm \sqrt{(0)^2-4(1)(-30)}}{2(1)}\] Simplify the expression inside the square root.

Step 4: Simplify the expression inside the square root

\[x = \frac{0 \pm \sqrt{0-(-120)}}{2}\] Since \(0-(-120)\) is equal to \(0 + 120\), we have: \[x = \frac{0 \pm \sqrt{120}}{2}\]

Step 5: Solve for \(x\)

Now, we have two solutions: \[x = \frac{\sqrt{120}}{2}\] \text{and} \[x = \frac{-\sqrt{120}}{2}\] Since \(\sqrt{120} = 2\sqrt{30}\), we simplify this to: \[x = \sqrt{30}\] \text{and} \[-\sqrt{30}\]

Key Concepts

quadratic equation

A quadratic equation is a second-degree polynomial equation in a single variable. Its general form is: \[ax^2 + bx + c = 0\] Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The coefficient \(a\) cannot be zero because it would then be a linear equation instead of quadratic. The solutions to a quadratic equation represent the points where the parabola intersects the x-axis. These solutions could be real or complex numbers.

solving equations

Solving a quadratic equation involves finding the values of \(x\) that make the equation true. One of the most commonly used methods is the quadratic formula, which is particularly useful when factoring is not straightforward. The quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]You can also solve quadratic equations by factoring (when possible), completing the square, or using graphing techniques. Each method has its advantages depending on the complexity of the equation. Always begin by rearranging the equation into its standard form before selecting a solution method.

simplifying expressions

Simplifying expressions is essential in solving equations as it helps make the problem more manageable. It often involves combining like terms, reducing fractions, and simplifying radicals. For example, in the given problem, after substituting the values into the quadratic formula, you simplify the expression inside the square root: \[x = \frac{0 \pm \sqrt{0 - (-120)}}{2}\]This simplifies to: \[x = \frac{0 \pm \sqrt{120}}{2}\]Further simplification converts \(\sqrt{120}\) to \(2\sqrt{30}\), leading to: \(x = \sqrt{30}\) and \(-\sqrt{30}\). Always aim to simplify the expressions as much as possible to easily interpret your solutions.

standard form of a quadratic equation

The standard form of a quadratic equation is crucial because many solution methods, including the quadratic formula, rely on it. The standard form is: \[ax^2 + bx + c = 0\]Here, \(a\), \(b\), and \(c\) are constants. For instance, in the exercise \(x^2 - 30 = 0\), we rearrange it into the standard form: \[x^2 + 0x - 30 = 0\]This makes it clear that \(a = 1\), \(b = 0\), and \(c = -30\). Working in the standard form ensures that you can apply any of the solution methods correctly and consistently.