Matemática Elementar

Problema 1

Simplify the rational expression. State any restrictions on the variable.

$\frac{x^2 - 9}{x^2 + 5x + 6}$

Problemas

1. Simplify the rational expression. State any restrictions on the variable.

$\frac{x^2 - 9}{x^2 + 5x + 6}$

2. Perform the indicated operation and simplify.

a) $\dfrac{2x}{x + 3} + \dfrac{5}{x - 1}$

b) $\dfrac{x^2 - 4}{x + 1} \cdot \dfrac{x + 1}{x - 2}$

3. Rewrite the expression using long division (or polynomial division):

$\frac{2x^3 + 3x^2 - 5x + 7}{x + 2}$

Write the result in the form $q(x) + \dfrac{r}{x + 2}$ , where $q(x)$ is the quotient and $r$ is the remainder.

4. Solve for $x$ and check for extraneous solutions:

$\frac{3}{x - 2} + \frac{1}{x + 1} = \frac{7}{(x - 2)(x + 1)}$

5. Solve for $x$ and check for extraneous solutions:

$\frac{x}{x - 3} = \frac{3}{x - 3} + 2$

Gabarito — Rational Functions

1. Factor numerator and denominator:

$\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x - 3)(x + 3)}{(x + 2)(x + 3)}$

Cancel the common factor $(x + 3)$ :

$= \frac{x - 3}{x + 2}$

Restrictions: $x \neq -3$ and $x \neq -2$ (values that make the original denominator zero).

2. a) The LCD is $(x + 3)(x - 1)$ :

$\frac{2x}{x + 3} + \frac{5}{x - 1} = \frac{2x(x - 1)}{(x + 3)(x - 1)} + \frac{5(x + 3)}{(x + 3)(x - 1)}$

$= \frac{2x^2 - 2x + 5x + 15}{(x + 3)(x - 1)} = \frac{2x^2 + 3x + 15}{(x + 3)(x - 1)}$

The numerator $2x^2 + 3x + 15$ has discriminant $9 - 120 = -111 < 0$ , so it does not factor over the reals.

Restrictions: $x \neq -3$ , $x \neq 1$ .

b) Factor and multiply:

$\frac{x^2 - 4}{x + 1} \cdot \frac{x + 1}{x - 2} = \frac{(x - 2)(x + 2)}{x + 1} \cdot \frac{x + 1}{x - 2}$

Cancel $(x + 1)$ and $(x - 2)$ :

$= x + 2$

Restrictions: $x \neq -1$ , $x \neq 2$ .

3. Divide $2x^3 + 3x^2 - 5x + 7$ by $x + 2$ using long division:

Step 1: $2x^3 \div x = 2x^2$ . Multiply: $2x^2(x + 2) = 2x^3 + 4x^2$ . Subtract:

$(2x^3 + 3x^2) - (2x^3 + 4x^2) = -x^2$

Bring down $-5x$ : $-x^2 - 5x$ .

Step 2: $-x^2 \div x = -x$ . Multiply: $-x(x + 2) = -x^2 - 2x$ . Subtract:

$(-x^2 - 5x) - (-x^2 - 2x) = -3x$

Bring down $+7$ : $-3x + 7$ .

Step 3: $-3x \div x = -3$ . Multiply: $-3(x + 2) = -3x - 6$ . Subtract:

$(-3x + 7) - (-3x - 6) = 13$

Result:

$\frac{2x^3 + 3x^2 - 5x + 7}{x + 2} = 2x^2 - x - 3 + \frac{13}{x + 2}$

Verification: $(2x^2 - x - 3)(x + 2) + 13 = 2x^3 + 4x^2 - x^2 - 2x - 3x - 6 + 13 = 2x^3 + 3x^2 - 5x + 7$ $\checkmark$

4. The LCD is $(x - 2)(x + 1)$ . Multiply both sides by $(x - 2)(x + 1)$ :

$3(x + 1) + 1(x - 2) = 7$

$3x + 3 + x - 2 = 7$

$4x + 1 = 7$

$4x = 6$

$x = \frac{3}{2}$

Check for extraneous solutions: $x = \frac{3}{2}$ makes $x - 2 = -\frac{1}{2} \neq 0$ and $x + 1 = \frac{5}{2} \neq 0$ . Neither denominator is zero.

Verification:

$\frac{3}{\frac{3}{2} - 2} + \frac{1}{\frac{3}{2} + 1} = \frac{3}{-\frac{1}{2}} + \frac{1}{\frac{5}{2}} = -6 + \frac{2}{5} = \frac{-30 + 2}{5} = \frac{-28}{5}$

$\frac{7}{(\frac{3}{2} - 2)(\frac{3}{2} + 1)} = \frac{7}{(-\frac{1}{2})(\frac{5}{2})} = \frac{7}{-\frac{5}{4}} = -\frac{28}{5}$

Both sides equal $-\frac{28}{5}$ . $\checkmark$

The solution is $x = \dfrac{3}{2}$ .

5. The LCD is $(x - 3)$ . Multiply both sides by $(x - 3)$ :

$x = 3 + 2(x - 3)$

$x = 3 + 2x - 6$

$x = 2x - 3$

$-x = -3$

$x = 3$

Check for extraneous solutions: $x = 3$ makes the denominators $x - 3 = 0$ . Division by zero is undefined.

Therefore $x = 3$ is extraneous, and the equation has no solution.

This problem illustrates why checking for extraneous solutions is essential when solving rational equations — the algebraic process can produce "solutions" that are not valid in the original equation.