Question10. $x+\frac{6}{x}=-7$
12. $2-\frac{3}{x+4}=\frac{12}{x^{2}+4 x}$

Studdy Solution
Solve the quadratic equation using the quadratic formula, factoring, or completing the square. Here, we factor:
$$(x + 6)(x + 1) = 0$$
Set each factor equal to zero:
$$x + 6 = 0 \quad \text{or} \quad x + 1 = 0$$
Solve for $x$:
$$x = -6 \quad \text{or} \quad x = -1$$
The solutions for problem 10 are:
$$x = -6 \quad \text{and} \quad x = -1$$
**Problem 12:**
STEP_1: Factor the denominator on the right-hand side:
$$x^2 + 4x = x(x + 4)$$
Thus, the equation becomes:
$$2 - \frac{3}{x+4} = \frac{12}{x(x+4)}$$
STEP_2: Find a common denominator for the entire equation, which is $x(x+4)$:
Multiply every term by $x(x+4)$:
$$2x(x+4) - 3x = 12$$
Simplify:
$$2x^2 + 8x - 3x = 12$$ $$2x^2 + 5x = 12$$
STEP_3: Rearrange the equation into a standard quadratic form:
$$2x^2 + 5x - 12 = 0$$
Solve the quadratic equation using the quadratic formula, factoring, or completing the square. Here, we use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 2$, $b = 5$, and $c = -12$.
Calculate the discriminant:
$$b^2 - 4ac = 5^2 - 4(2)(-12) = 25 + 96 = 121$$
Calculate $x$:
$$x = \frac{-5 \pm \sqrt{121}}{4}$$ $$x = \frac{-5 \pm 11}{4}$$
This gives two solutions:
$$x = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad x = \frac{-16}{4} = -4$$
The solutions for problem 12 are:
$$x = \frac{3}{2} \quad \text{and} \quad x = -4$$