Problem 16 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\frac{4 x+3}{x+1}+\frac{2}{x}=\frac{1}{x^{2}+x}$$

Step-by-Step Solution

Answer

The only valid solution is \(-\frac{1}{4}\).

Step 1 - Identify a common denominator

First, to solve the equation \(\frac{4 x+3}{x+1}+\frac{2}{x}=\frac{1}{x^{2}+x}\), note the common denominator can be \(x(x+1)\).

Step 2 - Rewrite each fraction with the common denominator

Rewrite each fraction with the common denominator \(x(x+1)\): \(\frac{4x+3}{x+1} = \frac{(4x+3)x}{x(x+1)} = \frac{4x^2+3x}{x(x+1)}\), \(\frac{2}{x} = \frac{2(x+1)}{x(x+1)} = \frac{2x+2}{x(x+1)}\), and \(\frac{1}{x^2+x} = \frac{1}{x(x+1)}\).

Step 3 - Combine the fractions on the left side

Add the fractions on the left side: \(\frac{4x^2 + 3x}{x(x+1)} + \frac{2x + 2}{x(x+1)} = \frac{4x^2 + 3x + 2x + 2}{x(x+1)} = \frac{4x^2 + 5x + 2}{x(x+1)}\).

Step 4 - Set the numerators equal

With the same denominator on both sides, set the numerators equal: \(4x^2 + 5x + 2 = 1\).

Step 5 - Rearrange into a standard quadratic equation

Subtract 1 from both sides to form a standard quadratic equation: \(4x^2 + 5x + 1 = 0\).

Step 6 - Solve the quadratic equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = 5\), and \(c = 1\) to find the roots. Calculate the discriminant first: \(b^2 - 4ac = 5^2 - 4 \times 4 \times 1 = 25 - 16 = 9\).

Step 7 - Calculate the roots

Now solve for \(x\): \(x = \frac{-5 \pm \sqrt{9}}{8} = \frac{-5 \pm 3}{8}\). This yields two solutions: \(x = \frac{-5 + 3}{8} = \frac{-2}{8} = -\frac{1}{4}\) and \(x = \frac{-5 - 3}{8} = \frac{-8}{8} = -1\).

Step 8 - Verify the solutions

Check if both \(-\frac{1}{4}\) and \(-1\) satisfy the original equation. Notice that \(x = -1\) causes a zero denominator in the original equation, so it is not a valid solution. Therefore, the only valid solution is \(-\frac{1}{4}\).

Key Concepts

quadratic equation

A quadratic equation is any equation that can be rewritten in the standard form: \[ax^2 + bx + c = 0\]Here,

• \(a\) is the coefficient of \(x^2\) (the quadratic term)
• \(b\) is the coefficient of \(x\) (the linear term)
• \(c\) is the constant term

The goal when solving a quadratic equation is to find the values of \(x\) that make the equation true. These values are called the roots or solutions of the quadratic equation. In the given exercise, after combining and simplifying the fractions, we arrived at the quadratic equation \(4x^2 + 5x + 1 = 0\).

common denominator

When dealing with the addition or subtraction of fractions, it is important to have a common denominator among all fractions involved. A common denominator is a shared multiple of the denominators of the individual fractions.

In the exercise, the original fractions have the denominators \(x+1\), \(x\), and \(x^2+x\). Finding a common denominator allows these fractions to be combined into a single rational expression, simplifying the equation. By examining the denominators, we recognize that \(x(x+1)\) is the least common multiple.

• For example, \(\frac{4x+3}{x+1}\) needs to be rewritten with the denominator \(x(x+1)\)
• This becomes \(\frac{(4x+3)x}{x(x+1)} = \frac{4x^2+3x}{x(x+1)}\).
• Similarly, \(\frac{2}{x}\) becomes \(\frac{2(x+1)}{x(x+1)} = \frac{2x+2}{x(x+1)}\).

fraction addition

Once all fractions have the same denominator, you can add or subtract them much like regular numbers, but focusing only on their numerators.

In the given problem, after rewriting each fraction with the common denominator \(x(x+1)\), we get:

• \(\frac{4x^2 + 3x}{x(x+1)} + \frac{2x + 2}{x(x+1)} = \frac{4x^2 + 3x + 2x + 2}{x(x+1)} = \frac{4x^2 + 5x + 2}{x(x+1)}\)

Now we combined the two fractions into one. Since the denominators are equal, we can compare the numerators directly to simplify the combined fraction.

quadratic formula

To solve a quadratic equation of the form \(ax^2 + bx + c = 0\), we use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. First, we calculate the discriminant \(b^2 - 4ac\) which helps determine the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If zero, there is one real root.
• If negative, the roots are complex (imaginary).

In our example, for \(4x^2 + 5x + 1 = 0\):

• \(a = 4\)
• \(b = 5\)
• \(c = 1\)
We then compute the discriminant: \(5^2 - 4 \times 4 \times 1 = 25 - 16 = 9\).
Finally, by substituting \(a\), \(b\), and \(c\) back into the quadratic formula, we find the roots \( x = \frac{-5 \pm \sqrt{9}}{8}\). This results in the solutions \(x = -1\)
and \(x = -\frac{1}{4}\).
However, only \(x = -\frac{1}{4}\) is valid because \(x = -1\) yields a zero denominator in the original equation.