Problem 69 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ \left(x+\frac{1}{4}\right)^{2}=\frac{3}{16} $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\sqrt{3} - 1}{4} \) and \( x = \frac{-\sqrt{3} - 1}{4}.\)

Step 1 - Recognize the equation format

Identify that the given equation \(\left(x+\frac{1}{4}\right)^{2}=\frac{3}{16}\) is in the format \((a+b)^2=c.\)

Step 2 - Apply the square root property

To solve for \(x\), take the square root of both sides of the equation \(\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{3}{16}}.\) This simplifies to \( \left| x + \frac{1}{4} \right| = \frac{\sqrt{3}}{4} \).

Step 3 - Remove the absolute value

The absolute value equation \(\left| x + \frac{1}{4} \right| = \frac{\sqrt{3}}{4}\) leads to two separate equations: \(x + \frac{1}{4} = \frac{\sqrt{3}}{4}\) and \(x + \frac{1}{4} = -\frac{\sqrt{3}}{4}.\)

Step 4 - Solve the first equation

For \(x + \frac{1}{4} = \frac{\sqrt{3}}{4},\) subtract \( \frac{1}{4} \) from both sides to get \( x = \frac{\sqrt{3}}{4} - \frac{1}{4} = \frac{\sqrt{3} - 1}{4}.\)

Step 5 - Solve the second equation

For \(x + \frac{1}{4} = -\frac{\sqrt{3}}{4},\) subtract \( \frac{1}{4} \) from both sides to get \(x = -\frac{\sqrt{3}}{4} - \frac{1}{4} = \frac{-\sqrt{3} - 1}{4}.\)

Step 6 - State the solutions

The solutions to the equation \( \left(x+\frac{1}{4}\right)^{2}=\frac{3}{16} \) are \(x = \frac{\sqrt{3} - 1}{4} \) and \(x = \frac{-\sqrt{3} - 1}{4}.\)

Key Concepts

square root property

The square root property is a technique used to solve quadratic equations of the form \((a + b)^2 = c \). This property states that if \((x - p)^2 = n \), then the solution to the equation is \[ |x - p| = \sqrt{n} \]. By taking the square root of both sides, we eliminate the square by converting the squared term into an absolute value.
This results in two linear equations to solve for \( x \). It's important to ensure the radical on the right side is simplified completely.

algebraic equations

Algebraic equations are mathematical statements that contain variables and constants. Solving them involves finding the value(s) of the variable(s) that make the equation true. For example, in our exercise \( (x + \frac{1}{4})^2 = \frac{3}{16} \), our goal is to isolate \ x \ and solve for it.
The approach often includes various operations such as addition, subtraction, multiplication, and division. Breaking the problem into smaller steps and methodically applying algebraic rules simplifies the process.

absolute value equations

Absolute value equations involve expressions within absolute value notation \(|x| \). An absolute value equation \ |A| = B\ has two possible solutions: \A = B\ and \A = -B\ because each solution satisfies the assertion that \ x \ is \ B \ units away from zero.
In our exercise, after taking the square root of both sides, we get \|x + \frac{1}{4}| = \frac{\sqrt{3}}{4}.\ We then solve for \ x \ by setting up two separate linear equations: \ x + \frac{1}{4} = \frac{\sqrt{3}}{4} \ and \ x + \frac{1}{4} = -\frac{\sqrt{3}}{4} \ and solve each individually.

radical simplification

Radical simplification involves simplifying expressions that contain square roots or other roots. Simplifying radicals makes the solution cleaner and more manageable. \(\sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{4} \) is an example of radical simplification within our problem.
This involves simplifying both the numerator and the denominator to their simplest form. Regular practice helps in mastering these simplifications, which then become a powerful tool in solving quadratic and algebraic equations efficiently.