Problem 57 - Beginning and Intermediate Algebra

Question

Solve each equation by completing the square. Give (a) exact solutions and (b) solutions rounded to the nearest thousandth. \(3 r^{2}-2=6 r+3\)

Step-by-Step Solution

Answer

Exact: \(1 \pm \frac{2\sqrt{6}}{3}\); Approximate: 2.633, -0.633.

Step 1 - Reorganize the equation

Start by moving all terms to one side of the equation: \[ 3r^2 - 6r - 5 = 0 \].

Step 2 - Divide by the leading coefficient

Divide through by the coefficient of the quadratic term (3): \[ r^2 - 2r - \frac{5}{3} = 0 \].

Step 3 - Isolate the quadratic and linear terms

Move the constant term to the other side of the equation: \[ r^2 - 2r = \frac{5}{3} \].

Step 4 - Complete the square

To complete the square, add \( \left( \frac{b}{2} \right)^2 \) to both sides of the equation. Here, \( b = -2 \): \( \left( \frac{-2}{2} \right)^2 = 1 \). Add 1 to both sides: \[ r^2 - 2r + 1 = \frac{5}{3} + 1 \]. This becomes \[ r^2 - 2r + 1 = \frac{8}{3} \].

Step 5 - Factorize the perfect square trinomial

Rewrite the left-hand side as a squared binomial: \[ (r - 1)^2 = \frac{8}{3} \].

Step 6 - Solve for r

Take the square root of both sides: \[ r - 1 = \pm \sqrt{\frac{8}{3}} \]. This simplifies to \[ r - 1 = \pm \frac{2\sqrt{2}}{\sqrt{3}} \]. Rationalize the denominator: \[ r - 1 = \pm \frac{2\sqrt{6}}{3} \].

Step 7 - Isolate r

Add 1 to both sides to solve for \( r \): \[ r = 1 \pm \frac{2\sqrt{6}}{3} \].

Step 8 - Exact solutions

The exact solutions are: \[ r = 1 + \frac{2\sqrt{6}}{3} \] and \[ r = 1 - \frac{2\sqrt{6}}{3} \].

Step 9 - Approximate solutions

Evaluate the approximate values using a calculator: \[ 1 + \frac{2\sqrt{6}}{3} \approx 2.633 \] and \[ 1 - \frac{2\sqrt{6}}{3} \approx -0.633 \].

Key Concepts

solving quadratic equations

Solving quadratic equations is a foundational skill in algebra. Quadratic equations are equations that can be written in the form: ax^2 + bx + c = 0,
where

• a
and
• b
are



constants while

• x
is the variable.

One common method to solve quadratic equations is by completing the square. This method involves rewriting the quadratic equation to have a perfect square trinomial, making it easier to solve. Here is a step-by-step guide to solve the equation using the completing the square method.

Following the steps from the solutions:

• 1. Reorganize the equation such that all terms are on one side.
• 2. Divide through by the leading coefficient (if necessary).
• 3. Isolate the quadratic and linear terms.

These steps help to set up the equation for completing the square.

Let's now move on to the next key concept: exact solutions.

exact solutions

Exact solutions provide the precise values of the roots of the quadratic equation. When solving an equation by completing the square, we derive exact solutions before approximating them. After rearranging the equation and isolating terms, you complete the square:

• Add the square of half the coefficient of x to both sides of the equation.
• Form a perfect square trinomial on one side.

Let's use our example from the solution:

• Starting with: \[ r^2 - 2r = \frac{5}{3} \]
• Add 1 to both sides to get: \[ r^2 - 2r + 1 = \frac{8}{3} \]
• Factorize the left-hand side: \[ (r - 1)^2 = \frac{8}{3} \]

After forming a perfect square, solve for the variable by taking the square root of both sides and isolating the variable:\[ r - 1 = \pm \sqrt{\frac{8}{3}} \]
This step will yield the exact solutions, as shown:

• \[ r = 1 + \frac{2\sqrt{6}}{3} \]
• \[ r = 1 - \frac{2\sqrt{6}}{3} \]

Exact solutions are beneficial when the problem requires precise answers. However, they can be complex due to the presence of radicals.

Let's proceed to understanding approximate solutions.

approximate solutions

Approximate solutions are often easier to work with, especially in practical applications where exact values may not be necessary. To get these, you need to evaluate the exact solutions using a calculator. For our exercise, the exact solutions are:\[ r = 1 + \frac{2\sqrt{6}}{3} \]and \[ r = 1 - \frac{2\sqrt{6}}{3} \]. Use a calculator to find the approximate values:

• \[ 1 + \frac{2\sqrt{6}}{3} \approx 2.633 \]
• \[ 1 - \frac{2\sqrt{6}}{3} \approx -0.633 \]

These approximations are often rounded to a specific decimal place, making them easier to communicate and use. In our example, the solutions are rounded to the nearest thousandth.

Using approximate solutions is beneficial in contexts where extreme precision is not required, such as in everyday problem-solving or initial estimations in scientific work.

In summary, both exact and approximate solutions have their unique importance.