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Magazine Chapter 1: Problem 24
\(5-60\) Find all real solutions of the equation. $$ \frac{x}{2 x+7}-\frac{x+1}{x+3}=1 $$
Short Answer
Expert verified
The real solutions are \( x = -\frac{7}{3} \) and \( x = -4 \).
Step by step solution
01
Find a Common Denominator
To combine the fractions \( \frac{x}{2x+7} \) and \( \frac{x+1}{x+3} \) into a single fraction, we need a common denominator. The common denominator of \( 2x+7 \) and \( x+3 \) is their product, \((2x+7)(x+3)\). Each fraction must be adjusted to have this common denominator.
02
Express Each Fraction with the Common Denominator
Multiply the numerator and denominator of \( \frac{x}{2x+7} \) by \( x+3 \) and \( \frac{x+1}{x+3} \) by \( 2x+7 \). The equation becomes: \[ \frac{x(x+3) - (x+1)(2x+7)}{(2x+7)(x+3)} = 1 \]
03
Simplify the Numerator
Expand and simplify the numerators: \( x(x+3) = x^2 + 3x \) \( (x+1)(2x+7) = 2x^2 + 7x + 2x + 7 = 2x^2 + 9x + 7 \) Thus, the equation becomes: \[ \frac{x^2 + 3x - (2x^2 + 9x + 7)}{(2x+7)(x+3)} = 1 \] Simplifying, we get: \[ x^2 + 3x - 2x^2 - 9x - 7 = -x^2 - 6x - 7\]
04
Create an Equation Without Fractions
To eliminate the denominator, multiply both sides of the equation by \( (2x+7)(x+3) \): \[ -x^2 - 6x - 7 = (2x+7)(x+3) \]
05
Set Quadratic to Zero
Simplify the right side: \( (2x+7)(x+3) = 2x^2 + 7x + 6x + 21 = 2x^2 + 13x + 21 \) Now, the equation is: \[ -x^2 - 6x - 7 = 2x^2 + 13x + 21 \] Move all terms to one side to form a quadratic equation: \[ 0 = 3x^2 + 19x + 28 \] Or \[ 3x^2 + 19x + 28 = 0 \]
06
Solve the Quadratic Equation
To solve \( 3x^2 + 19x + 28 = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 19 \), and \( c = 28 \).Calculate the discriminant: \( b^2 - 4ac = 19^2 - 4 \cdot 3 \cdot 28 = 361 - 336 = 25 \)Then, find the solutions: \( x = \frac{-19 \pm \sqrt{25}}{6} \)There are two solutions:\( x = \frac{-19 + 5}{6} = \frac{-14}{6} = \frac{-7}{3} \) and \( x = \frac{-19 - 5}{6} = \frac{-24}{6} = -4 \).
07
Check for Restrictions
When solving rational equations, ensure the solutions don't make any denominator zero. For the original denominators \(2x+7\) and \(x+3\), solve for restrictions:- \( 2x+7 = 0 \) gives \( x = -\frac{7}{2} \) - \( x+3 = 0 \) gives \( x = -3 \)Neither \( x = -\frac{7}{3} \) nor \( x = -4 \) match these restrictions, so both are valid solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
In mathematics, a quadratic equation is a second-degree polynomial equation in the form of \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). This type of equation is crucial because it frequently appears in various mathematical contexts and real-world applications, such as physics and engineering.
To find the solutions of a quadratic equation, several methods can be employed:
To find the solutions of a quadratic equation, several methods can be employed:
- Factoring: Breaking down the equation into a product of simpler polynomials, possible when the equation is factorable.
- Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), applicable to any quadratic equation regardless of factorability.
- Completing the Square: Rewriting the equation in such a way that allows for extracting the square root, useful for understanding the derivation of the quadratic formula.
- Graphing: Identifying the points where the parabola crosses the x-axis, which are the solutions of the equation.
Common Denominator
The concept of a common denominator is crucial when working with fractions, especially in solving rational equations. A common denominator allows us to combine multiple fractions into a single expression by ensuring they all share the same base value.
In this exercise, we started with two fractions: \(\frac{x}{2x+7}\) and \(\frac{x+1}{x+3}\). To combine these, we looked for a common denominator, which is the product of both denominators in this case: \((2x+7)(x+3)\).
In this exercise, we started with two fractions: \(\frac{x}{2x+7}\) and \(\frac{x+1}{x+3}\). To combine these, we looked for a common denominator, which is the product of both denominators in this case: \((2x+7)(x+3)\).
- Adjust each fraction to the common denominator by multiplying the numerator and denominator by the necessary term from other fractions.
- For \(\frac{x}{2x+7}\), multiply by \(x+3\).
- For \(\frac{x+1}{x+3}\), multiply by \(2x+7\).
Discriminant
The discriminant is an important concept in algebra, particularly when dealing with quadratic equations. It is the part under the square root of the quadratic formula, denoted by \(b^2 - 4ac\). The discriminant helps us determine the nature and number of roots of a quadratic equation.
- Discriminant > 0: Two distinct real solutions. This indicates that the parabola crosses the x-axis in two different points.
- Discriminant = 0: One real solution. The parabola touches the x-axis at exactly one point. This scenario is also known as a perfect square trinomial.
- Discriminant < 0: No real solutions. The solutions are complex numbers, and the parabola does not intersect the x-axis.
