Question150 Solve: $$\frac{1}{4 r}=\frac{r^{2}+5 r+6}{r}-\frac{1}{2 r}$$

Studdy Solution
Set the left side equal to the simplified right side and solve for $r$:
$$\frac{1}{4r} = \frac{2r^2 + 10r + 11}{2r}$$
Cross-multiply to eliminate the fractions:
$$1 \cdot 2r = 4r(2r^2 + 10r + 11)$$
Simplify and solve the resulting equation:
$$2r = 8r^3 + 40r^2 + 44r$$
Rearrange the equation to set it to zero:
$$8r^3 + 40r^2 + 44r - 2r = 0$$ $$8r^3 + 40r^2 + 42r = 0$$
Factor out the greatest common factor, which is $2r$:
$$2r(4r^2 + 20r + 21) = 0$$
Set each factor equal to zero:
1. $2r = 0$ implies $r = 0$
2. Solve $4r^2 + 20r + 21 = 0$ using the quadratic formula:
$$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 4$, $b = 20$, and $c = 21$.
$$r = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 4 \cdot 21}}{2 \cdot 4}$$ $$r = \frac{-20 \pm \sqrt{400 - 336}}{8}$$ $$r = \frac{-20 \pm \sqrt{64}}{8}$$ $$r = \frac{-20 \pm 8}{8}$$
Calculate the roots:
$$r = \frac{-20 + 8}{8} = \frac{-12}{8} = -\frac{3}{2}$$ $$r = \frac{-20 - 8}{8} = \frac{-28}{8} = -\frac{7}{2}$$
The possible solutions for $r$ are $r = 0$, $r = -\frac{3}{2}$, and $r = -\frac{7}{2}$.
The solutions for $r$ are:
$$r = 0, \, r = -\frac{3}{2}, \, r = -\frac{7}{2}$$