Problem 32 - College Algebra

Question

Solve cach equation in Exercises \(15-34\) by the square root property. $$(8 x-3)^{2}=5$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x_1 = (3 + \sqrt{5}) / 8\) and \(x_2 = (3 - \sqrt{5}) / 8\)

Step 1: Isolate the Squared Expression

First, to observe the square root property, the squared expression needs to be alone on one side of the equation. In this case, the squared expression (8x-3)^2 is already alone and equals to 5, thus no step needed.

Step 2: Apply the Square Root Property

Then apply the square root property to both sides of the equation. By square root property, if (8x-3)^2 = 5, then 8x-3 must be equal to the positive or negative square root of 5. \(8x - 3 = \pm \sqrt{5}\). There are now two possible equations to solve.

Step 3: Solve for the Variable

Now simplify each equation by adding 3 to both sides and then dividing by 8, to isolate \(x\). The relevant equations are: \(x_1 = (3 + \sqrt{5}) / 8\) and \(x_2 = (3 - \sqrt{5}) / 8\)

Step 4: Simplify

The solutions \(x_1\) and \(x_2\) cannot be further simplified, hence these are the solutions to the equation.

Key Concepts

Solving Quadratic Equations

Quadratic equations are a fundamental part of algebra. They come in the form of \(ax^2 + bx + c = 0\) where \(a, b,\) and \(c\) are constants. The exercise employs a special scenario where we use the square root property to solve equations of the form \((ax + b)^2 = c\).
This means our quadratic doesn't start in its usual expanded form. Instead, it's compressed into a simple squared term. The goal is to unsquare this particular setup so that we can then find the values for the variable. This is a neat alternative to solving quadratics compared to factoring or using the quadratic formula.

Isolating the Variable

Isolating the variable means getting the variable, usually \(x\), by itself on one side of the equation. This process reduces the equation and simplifies solving it.
In our problem, the squared expression \((8x-3)^2 = 5\) is already isolated, meaning no additional steps were necessary initially. Once this expression has been resolved by unsquaring it, the new challenge is to work with each resulting equation from the positive and negative roots separately. This continues the isolation process, where we first add 3 to each side and then divide by 8. This way, our \(x\) is neatly isolated for each solution.

Radical Expressions

Radical expressions involve roots, such as square roots. When working with quadratic equations like our example, applying the square root property is critical. This property states that if \(a^2 = b\), then \(a = \pm \sqrt{b}\).
In applying this to our equation \((8x-3)^2 = 5\), this transformative step allows us to express \(8x - 3\) in terms of \(\pm \sqrt{5}\). This approach acknowledges that squaring a number gives a positive result, thus the roots must include both potential positive and negative outcomes, which doubles the solution possibilities.

Algebraic Solutions

Algebraic solutions emphasize using algebraic manipulations to arrive at the value(s) for the variable.
In our task, once we apply the square root property, we end up with two root equations: \(8x - 3 = \sqrt{5}\) and \(8x - 3 = -\sqrt{5}\). Solving these separately involves straightforward algebraic steps:

• Add 3 to each side of both equations.
• Then, divide the results by 8 to fully solve for \(x\).

In this way, algebraic solutions provide a satisfying method of breaking down the equation into manageable parts until producing the final answers \(x_1 = \frac{3 + \sqrt{5}}{8}\) and \(x_2 = \frac{3 - \sqrt{5}}{8}\). These showcase the methodical power of algebra in dissecting equations.