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Chapter 8: Problem 42

Solve. $$9\left(\frac{x+2}{x+3}\right)^{2}-6\left(\frac{x+2}{x+3}\right)+1=0$$

Short Answer

Expert verified
The solution to the equation is \( x = -\frac{3}{2} \).

Step by step solution

01

- Introduce substitution

Let us simplify the expression by introducing substitution. Let \( y = \frac{x+2}{x+3} \). Then the equation becomes: \[ 9y^2 - 6y + 1 = 0 \]
02

- Solve the quadratic equation

Now solve the quadratic equation \( 9y^2 - 6y + 1 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 9\), \(b = -6\) and \(c = 1\).
03

- Calculate discriminant

Calculate the discriminant: \[ \Delta = b^2 - 4ac \]\[ \Delta = (-6)^2 - 4(9)(1) = 36 - 36 = 0 \]
04

- Find roots of the quadratic equation

Since the discriminant is 0, there is one real root:\[ y = \frac{-(-6) \pm \sqrt{0}}{2(9)} \]\[ y = \frac{6}{18} \]\[ y = \frac{1}{3} \]
05

- Substitute back

Re-substitute \( y \) back to \( \frac{x+2}{x+3} \): \[ \frac{x+2}{x+3} = \frac{1}{3} \]
06

- Solve for x

Cross-multiply to solve for \( x \): \[ 3(x+2) = x + 3 \]\[ 3x + 6 = x + 3 \]\[ 2x = -3 \]\[ x = -\frac{3}{2} \]

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Key Concepts

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

Substitution Method
The **substitution method** is a powerful tool for simplifying complex equations. It involves replacing a complicated part of an equation with a single variable to make the equation easier to solve.
For example, in the original problem: \(9\frac{(x+2)^2}{(x+3)^2} -6 \frac{x+2}{x+3} +1 = 0\), we can let \(y = \frac{x+2}{x+3}\).
This substitution simplifies the original equation to \(9y^2 - 6y + 1 = 0\).
Now, instead of working with a rational expression, we can solve a simpler quadratic equation.
Once we solve for **y**, we can substitute back to find **x**. This method can turn seemingly difficult problems into manageable ones.
Quadratic Formula
The **quadratic formula** is essential for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.}\]
In our simplified equation from step 2: \(9y^2 - 6y + 1 = 0\), we substitute \(a = 9\), \(b = -6\), and \(c = 1\) into the quadratic formula.
This helps us to find the value(s) of **y** that satisfy the equation.
Knowing this formula allows us to solve any quadratic equation by simply plugging in the coefficients **a**, **b**, and **c**.
Discriminant
The **discriminant** is part of the quadratic formula under the square root: \(b^2 - 4ac\). It tells us about the nature of the solutions of the quadratic equation:
  • If the discriminant is positive, the equation has two distinct real roots.
  • If it is zero, there is exactly one real root.
  • If it is negative, there are no real roots (but two complex roots).

In our problem, the discriminant is calculated as follows: \(b^2 - 4ac = (-6)^2 - 4(9)(1) = 36 - 36 = 0\).
Since the discriminant is zero, our equation \(9y^2 - 6y + 1 = 0\) has exactly one real root.
Solving Quadratic Equations
In general, **solving quadratic equations** involves several steps:
  • First, express the equation in the standard quadratic form \(ax^2 + bx + c = 0\).
  • Check if the equation can be factored easily. If not, use the quadratic formula.
  • Calculate the discriminant \(b^2 - 4ac\).
  • Apply the quadratic formula to find the solutions for the variable.

Let's revisit the problem after our substitution and simplification.
We found the value of **y** as \(y = \frac{1}{3}\). By substituting back into \(y = \frac{x+2}{x+3}\), we solved for **x** and found \(x = -\frac{3}{2}\).
Notice how breaking down the problem into smaller steps makes it easier to manage and solve.

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