Problem 51 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ (x-4)^{2}=3 $$

Step-by-Step Solution

Answer

x = 4 \pm \sqrt{3}

Step 1: Isolate the squared term

The equation is already in the form \( (x-4)^{2}=3 \), where the squared term \( (x-4)^{2} \) is isolated.

Step 2: Apply the square root property

Take the square root of both sides of the equation: \[ \sqrt{(x-4)^{2}} = \sqrt{3} \]. This simplifies to \( x-4 = \pm \sqrt{3} \).

Step 3: Solve for x

Set up two equations using \( \pm \sqrt{3} \): \[ x - 4 = \sqrt{3} \] and \[ x - 4 = - \sqrt{3} \].

Step 4: Solve each equation

Add 4 to both sides of each equation to isolate x: \[ x = 4 + \sqrt{3} \] and \[ x = 4 - \sqrt{3} \].

Key Concepts

square root property

To solve the given equation \((x-4)^{2}=3\) using the square root property, start by isolating the term that is squared. We already have the isolated squared term \((x-4)^{2}\).
Next, apply the square root property by taking the square root of both sides of the equation. This means we will write: \(\sqrt{(x-4)^{2}} = \sqrt{3}\).
According to the square root property, the square root of a squared term \((a^2)\) equals the absolute value of that term, which means these can be simplified to: \(x - 4 = \pm \sqrt{3}\).
The \(\pm\) sign indicates that there are two possible solutions: one is positive \(\sqrt{3}\) and the other is negative \(\-\sqrt{3}\).
Remember that this property works not just for simple numbers but for more complex expressions as well, which is very useful.

simplifying radicals

When we talk about simplifying radicals, we are looking to express the radical in its simplest form. In this equation, we have \(\sqrt{3}\).
Radical simplification involves factors of numbers. For example, if you have a radical like \(\sqrt{12}\), you would break it down into its factors \(\sqrt{4 \cdot 3}\), which simplifies to \(2\sqrt{3}\). However, \(\sqrt{3}\) is already in its simplest form because 3 is a prime number.
Simplifying radicals becomes particularly important when dealing with larger numbers or multiple terms under the radical sign.
So in our problem, \(\sqrt{3}\) remains as it is, and there's no further simplification needed for this particular example.

isolating variables

The process of isolating variables involves getting the variable you are solving for, on one side of the equation by itself. In our example, we need to isolate \(x\).
After applying the square root property, we have two equations: \(x - 4 = \sqrt{3}\) and \(x - 4 = -\sqrt{3}\).
To isolate \(x\) from each equation, we need to add 4 to both sides in both equations.
For \(x - 4 = \sqrt{3}\), adding 4 to both sides gives: \(x = 4 + \sqrt{3}\).
For \(x - 4 = -\sqrt{3}\), adding 4 to both sides gives: \(x = 4 - \sqrt{3}\).
Now the variable \(x\) stands alone, and we have our solutions: \(x = 4 + \sqrt{3}\) and \(x = 4 - \sqrt{3}\).
Isolating the variable is a crucial step that helps to clearly see the solutions for the equation.