Problem 34 - College Algebra Essentials

Question

Solve each equation in Exercises \(15-34\) by the square root property. $$ (2 x+8)^{2}=27 $$

Step-by-Step Solution

Answer

The solution for x in the given equation is \(x = \frac {-8 \pm \sqrt{27}}{2}\)

Step 1: Isolate the square

Start by isolating the square on one side of the equation. Here, the square is on the left hand side already hence nothing to do here.

Step 2: Apply the square root property

We are now ready to apply the square root property. Applying square root on both sides: \(\sqrt{(2x+8)^2} = \pm \sqrt {27}\). This leaves us with: \(2x+8 = \pm \sqrt {27}\)

Step 3: Decrease the effect of constant

Decreate the effect of constant from the left side by subtracting 8 from both sides: \(2x = \pm \sqrt {27} - 8.\)

Step 4: Decrease the effect of coefficient

Finally, decrease the effect of coefficient of x, which is 2 in this case, by dividing both sides by 2: \(x = \frac {\pm \sqrt {27} - 8}{2}.\)

Key Concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is a fundamental skill in algebra. These equations can often be recognized by their standard form, which is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), are coefficients and \( a \) is not equal to zero. To solve such equations, you can use various methods such as factoring, completing the square, using the quadratic formula, or the square root property, as in the provided textbook exercise.

When the quadratic expression is a perfect square trinomial, the square root property becomes particularly handy. This property states that if \( x^2 = k \), where \( k \), is a non-negative real number, then \( x \) is equal to \( \pm\sqrt{k} \). But remember, when you take the square root of both sides of an equation, always consider both the positive and negative square roots to find all solutions to the equation.

Isolate the Square

The process of isolating the square involves manipulating the equation so that the term with an exponent of two, the square term, stands alone on one side of the equation. This is typically the first step in using the square root property effectively.

To isolate the square, you might need to perform operations such as adding or subtracting terms to both sides of the equation, or dividing by a coefficient. In our exercise example, the square term \( (2x+8)^2 \) is already on one side of the equation, so no further action is required to isolate it. However, suppose we had an equation with additional terms on the same side as the square, you would need to move those terms to the opposite side before applying the square root property.

Square Root Operations

Once the square has been isolated, we can apply the operations of taking square roots to solve the equation. The key is to understand that the square root of a squared term will yield the absolute value of the original term. In simpler terms, \( \sqrt{x^2} = |x| \), which represents both the positive and negative values of \( x \).

For instance, in our textbook solution \( \sqrt{(2x+8)^2} \), resolves into \( \pm(2x+8) \), reflecting both possible values of \( x \) that can satisfy the squared equation. It's essential to include the \( \pm \) symbol after taking the root to maintain the equation's integrity. Failing to do so can lead to missing one of the potential solutions.