Problem 77 - Elementary Algebra

Question

Solve and round off the solutions to the nearest hundredth. $$ (2 x+1)(x+3)-(x+7)=(x+3) 2 $$

Step-by-Step Solution

Answer

The solutions are approximately 1.61 and -3.11.

Step 1: Expand Both Sides

First, expand both sides of the equation. Start with the left side: \[ (2x+1)(x+3) - (x+7) = 2x(x+3) + 1(x+3) - x - 7 \]Now expand:\[ = 2x^2 + 6x + x + 3 - x - 7 \] Combine like terms:\[ = 2x^2 + 6x + 3 - 7 \] Simplify:\[ = 2x^2 + 5x - 4 \]Now for the right side:\[ (x+3)2 = 2(x+3) = 2x + 6 \] So the equation becomes:\[ 2x^2 + 5x - 4 = 2x + 6 \]

Step 2: Rearrange the Equation

Move all terms to one side of the equation to set it to zero:\[ 2x^2 + 5x - 4 - 2x - 6 = 0 \] Combine the terms:\[ 2x^2 + 3x - 10 = 0 \]

Step 3: Apply the Quadratic Formula

We now apply the quadratic formula to solve for \(x\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \(a = 2\), \(b = 3\), and \(c = -10\). Substitute these into the formula:\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2} \] Simplify under the square root:\[ x = \frac{-3 \pm \sqrt{9 + 80}}{4} \] \[ = \frac{-3 \pm \sqrt{89}}{4} \]

Step 4: Compute the Solutions

Calculate the solutions by evaluating the expression separately for both the plus and minus results from the quadratic formula:1. \(+\sqrt{89}: \) \[ x_1 = \frac{-3 + \sqrt{89}}{4} \approx \frac{-3 + 9.43}{4} \approx \frac{6.43}{4} \approx 1.61 \]2. \(-\sqrt{89}: \) \[ x_2 = \frac{-3 - \sqrt{89}}{4} \approx \frac{-3 - 9.43}{4} \approx \frac{-12.43}{4} \approx -3.11 \]

Step 5: Round to the Nearest Hundredth

The solutions calculated earlier were already rounded to the nearest hundredth. Thus, we have:\[ x_1 \approx 1.61 \]\[ x_2 \approx -3.11 \]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows you to find the roots or solutions of the equation. Here's how it works:

• Identify \( a \), \( b \), and \( c \): These come from the standard form of the quadratic equation.
• Calculate the discriminant: \( b^2 - 4ac \). The value under the square root symbol.
• Determine the solutions: Use the '+' and '-' signs to find two potential solutions.
• Solve: Evaluate each solution by plugging values back into the equation.

Using the quadratic formula guarantees we can solve any quadratic equation, even when factoring is not possible.

Algebraic Expansion

Algebraic expansion involves multiplying terms to remove parentheses in an expression. Consider the expression \((2x + 1)(x + 3)\). To expand it:

• Distribute each term: Multiply each term in the first parenthesis by each term in the second. For instance, \(2x \times x \), \(2x \times 3 \), \(1 \times x \), \(1 \times 3 \).
• Write out the results: Combine them all to form: \(2x^2 + 6x + x + 3\).
• Simplify: Combine like terms which gives us \(2x^2 + 7x + 3\).

This step is crucial in simplifying expressions and setting the foundation for solving equations. It makes it visually easier to work with and apply further algebraic operations.

Combining Like Terms

Combining like terms is a method used to simplify algebraic expressions. Terms are 'like' if they share the same variable to the same power. Consider the expression we expanded: \(2x^2 + 6x + x + 3\). To simplify it:

• Identify like terms: Notice that \(6x\) and \(x\) are like terms since both have the variable \(x\).
• Combine the coefficients: Add the numbers in front of the like terms, which yields \(6 + 1 = 7\), thus \(7x\).
• Rewrite the expression: \(2x^2 + 7x + 3\).

By combining like terms, the expression is easier to handle and see further steps in solving the equation. It is a fundamental skill in algebra that helps in reducing expressions to their simplest form.

Rounding Numbers

Rounding numbers is the process of adjusting values to a certain degree of precision, often to make them easier to work with. When asked to round a number to the nearest hundredth, you look at the digits after the decimal point.

• Locate the hundredth place: This is the second digit to the right of the decimal point.
• Check the digit after it: For instance, in the number 1.612, the hundredth place is '1'.
• Decide how to round: If the next digit is 5 or more, round up. So, 1.612 becomes 1.61.
• Adjust accordingly: Conversely, if the next digit is 4 or less, keep the number as is.

Rounding helps manage precision, especially useful in final presentations of solutions where exact decimal points are unnecessary. It creates a cleaner, more comprehensible result.