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Chapter 11: Problem 27

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ (r-3)(r+5)=2 $$

Short Answer

Expert verified
r = -1 + 3\sqrt{2} and r = -1 - 3\sqrt{2}

Step by step solution

01

Expand the Equation

First, expand the given equation \( (r-3)(r+5) = 2 \). Multiply the expressions \( (r-3) \) and \( (r+5) \).
02

Multiply the Binomials

Apply the distributive property: \[ (r-3)(r+5) = r^2 + 5r - 3r - 15 \]. Simplify to get: \[ r^2 + 2r - 15 = 2 \].
03

Move All Terms to One Side

Subtract 2 from both sides: \[ r^2 + 2r - 15 - 2 = 0 \], which simplifies to: \[ r^2 + 2r - 17 = 0 \].
04

Identify Quadratic Formula Components

The quadratic formula is: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Identify the coefficients from the equation \( r^2 + 2r - 17 = 0 \): \( a = 1, b = 2, \) and \( c = -17 \).
05

Substitute into Quadratic Formula

Substitute the values into the quadratic formula: \[ r = \frac{-2 \pm \sqrt{2^2 - 4(1)(-17)}}{2(1)} \]. Simplify inside the square root: \[ r = \frac{-2 \pm \sqrt{4 + 68}}{2} \], leading to \[ r = \frac{-2 \pm \sqrt{72}}{2} \].
06

Simplify the Expression

Simplify \( \sqrt{72} \): \[ \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \]. Substitute back in: \[ r = \frac{-2 \pm 6\sqrt{2}}{2} \]. Simplify the fractions: \[ r = -1 \pm 3\sqrt{2} \].
07

State the Solutions

The solutions to the equation are: \[ r = -1 + 3\sqrt{2} \] and \[ r = -1 - 3\sqrt{2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expanding Equations
To solve equations like \( (r-3)(r+5) = 2 \), we first need to expand the equation. Expanding involves removing parentheses and applying the distributive property, allowing us to rewrite the equation in a simpler form. Here, we start by expanding the binomials \( (r-3)(r+5) \). This results in combining all terms together.
Distributive Property
The distributive property is fundamental in simplifying expressions and equations. It states that \ a(b + c) = ab + ac \. Using this property for \( (r-3)(r+5) \), distribute each term in the binomial separately: \( r \cdot \ r + r \cdot \ 5 - 3 \cdot \ r - 3 \cdot \ 5 = r^2 + 5r - 3r - 15 \). Combine like terms to simplify further.
Quadratic Equations
After distributing and combining like terms, our equation \( r^2 + 2r - 15 = 2 \) becomes a quadratic equation. Quadratic equations are polynomials of degree 2, typically written as \( ax^2 + bx + c = 0 \). In our case, subtracting 2 from both sides gives us: \( r^2 + 2r - 17 = 0 \), where \ a = 1, b = 2, \ and \ c = -17 \.
Solving Quadratic Equations
To solve quadratic equations, we use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Identify the coefficients and substitute into the formula: \[ r = \frac{-2 \pm \sqrt{2^2 - 4\cdot1\cdot(-17)}}{2\cdot1} \] which simplifies to \[ r = \frac{-2 \pm \sqrt{4 + 68}}{2} = \frac{-2 \pm \sqrt{72}}{2} \]. Simplify \sqrt{72} = 6\sqrt{2} \, resulting in the final solutions: \[ r = -1 + 3\sqrt{2} \] and \[ r = -1 - 3\sqrt{2} \].

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