SOLUTION: Find all values of $a$ that satisfy the equation \frac{a}{4} + 1 = \frac{a + 4}{a^2} + 7.

Question 1209052: Find all values of $a$ that satisfy the equation
\frac{a}{4} + 1 = \frac{a + 4}{a^2} + 7.
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
To solve the equation, we'll first clear the denominators by multiplying both sides by $4a^2$:
$$a^3 + 4a^2 = 4(a + 4) + 28a^2$$
Expanding and rearranging the terms, we get:
$$a^3 - 24a^2 - 16a - 16 = 0$$
Unfortunately, there's no simple algebraic method to solve cubic equations in general. We can use numerical methods or software tools to find approximate solutions. However, we can try to factor the equation to see if we can find any rational solutions.
By trial and error, we can find that $a = 4$ is a solution. This means that $(a-4)$ is a factor of the cubic polynomial. We can perform polynomial long division to find the other factor:
$$a^3 - 24a^2 - 16a - 16 = (a-4)(a^2 - 20a - 4)$$
The quadratic factor $a^2 - 20a - 4$ can be solved using the quadratic formula:
$$a = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(-4)}}{2(1)}$$
Simplifying, we get:
$$a = 10 \pm 2\sqrt{26}$$
Therefore, the solutions to the original equation are:
$$a = 4, 10 + 2\sqrt{26}, 10 - 2\sqrt{26}$$

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