Problem 137 - Elementary Algebra

Question

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve. (a) \(6 a^{2}+14=20\) (b) \(\left(x-\frac{1}{4}\right)^{2}=\frac{5}{16}\) (c) \(y^{2}-2 y=8\)

Step-by-Step Solution

Answer

(a) Factoring (b) Square Root (c) Quadratic Formula

Step 1 - Identify the method for equation (a)

Given the equation: \[6a^2 + 14 = 20\] First, simplify the equation: \[6a^2 + 14 - 20 = 0\] Simplifying results in: \[6a^2 - 6 = 0\] Factor out the common term: \[6(a^2 - 1) = 0\] Since the equation can be factored easily, Factoring is the most appropriate method.

Step 2 - Identify the method for equation (b)

Given the equation: \[\bigg(x - \frac{1}{4}\bigg)^2 = \frac{5}{16}\] This equation is already in a form that suggests taking the square root of both sides. Thus, the most appropriate method to solve this equation is the Square Root method.

Step 3 - Identify the method for equation (c)

Given the equation: \[y^2 - 2y = 8\] First, rewrite the equation in standard form: \[y^2 - 2y - 8 = 0\] This quadratic equation can be either factored or solved using the Quadratic Formula. If factoring is not straightforward, the Quadratic Formula is a reliable method. Since factoring might not be immediate, the Quadratic Formula is the most appropriate method.

Key Concepts

factoring quadratic equations

When you have a quadratic equation, one effective method to solve it is by factoring. Factoring involves breaking down the quadratic expression into a product of simpler expressions that can be set to zero. The general form of a quadratic equation is expressed as \(ax^2 + bx + c = 0\).
Here are the steps to factor a quadratic equation:

• Identify the coefficients \(a\), \(b\), and \(c\) in the quadratic equation \(ax^2 + bx + c\).
• Find two numbers that multiply to \(a \times c\) and add up to \(b\).
• Rewrite the middle term \(bx\) using the two numbers found in the previous step.
• Factor by grouping, which involves pairing terms from the expanded equation.
• Simplify and solve for the variable by setting each factor equal to zero.

For example, consider \(6a^2 - 6 = 0\). Factoring out the common term \(6\) gives us \(6(a^2 - 1) = 0\). Further, \(a^2 - 1\) can be factored into \((a - 1)(a + 1)\). So, the solutions are \(a = 1\) and \(a = -1\).
Factoring is often straightforward when the coefficients are simple and easily manipulable fractions, but it might require practice to recognize factoring opportunities in more complex equations.

square root method

The square root method is ideal when a quadratic equation can be expressed in the form \((x - p)^2 = q\). This method simplifies solving by directly taking the square root of both sides. Here’s how:

• Isolate the squared term on one side of the equation.
• Take the square root of both sides of the equation.
• Solve for the variable by simplifying and handling any constants.

For instance, with the equation \(\left(x - \frac{1}{4}\right)^2 = \frac{5}{16}\), you take the square root of both sides to get \(x - \frac{1}{4} = \pm \frac{\sqrt{5}}{4}\). Then, solve for \(x\) by adding \(\frac{1}{4}\) to both sides.
The square root method is typically used when the equation is already in a quadratic perfect square form. It’s a straightforward and efficient technique for solving special cases of quadratic equations.

quadratic formula

The quadratic formula is a universal tool for solving any quadratic equation! Whether the equation can be factored or not, this formula will always work. The quadratic formula is:

• Given the standard form \(ax^2 + bx + c = 0\), identify \(a\), \(b\), and \(c\).
• Apply the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• Simplify under the square root (the discriminant), then divide by \(2a\) to get the solutions.

Take, for example, the equation \(y^2 - 2y - 8 = 0\). This translates to \(a = 1\), \(b = -2\), and \(c = -8\). Substituting these into the quadratic formula yields: \[y = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]
This provides the solutions: \(y = 4\) and \(y = -2\).
Using the quadratic formula ensures that no matter the complexity of the quadratic equation, you’ll find the roots accurately.

standard form of quadratic equations

Understanding the standard form of quadratic equations is crucial for applying various solving methods. The standard form is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Here's why it's useful:

• It standardizes the approach to solving, ensuring methods like factoring and the quadratic formula are directly applicable.
• Identifying the coefficients \(a\), \(b\), and \(c\) helps determine the most efficient solving method.
• Rewriting equations into standard form can simplify recognition of patterns like perfect squares or factorable numbers.

For instance, convert the equation \(y^2 - 2y = 8\) into standard form by subtracting \(8\) from both sides to get \(y^2 - 2y - 8 = 0\). Now, it’s clear this can be solved by factoring or through the quadratic formula.
Consistently putting quadratic equations in standard form aligns them for all major solving methods and makes analysis straightforward.