Solve for \( u \). \[ 3+\frac{1}{u+6}=-\frac{5}{(u+1)(u+6)} \] If there is more than one solution, separate them with If there is no solution, click on "No solution".
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To solve for \( u \) in the equation \[ 3+\frac{1}{u+6}=-\frac{5}{(u+1)(u+6)}, \] first, let's eliminate fractions by multiplying through by \((u+1)(u+6)\): \[ 3(u+1)(u+6) + (u+1) = -5. \] Expanding both sides: \[ 3(u^2 + 7u + 6) + u + 1 = -5. \] This simplifies to: \[ 3u^2 + 21u + 18 + u + 1 = -5. \] Combine like terms: \[ 3u^2 + 22u + 19 = -5. \] Now, add 5 to both sides: \[ 3u^2 + 22u + 24 = 0. \] Next, we can apply the quadratic formula where \( a = 3 \), \( b = 22 \), and \( c = 24 \): \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-22 \pm \sqrt{22^2 - 4 \cdot 3 \cdot 24}}{2 \cdot 3}. \] Calculating the discriminant: \[ 22^2 - 4 \cdot 3 \cdot 24 = 484 - 288 = 196. \] Since the discriminant is positive, we have two real solutions. Now plugging this back into the quadratic formula: \[ u = \frac{-22 \pm \sqrt{196}}{6} = \frac{-22 \pm 14}{6}. \] This gives us two possible values for \( u \): 1. \( u_1 = \frac{-22 + 14}{6} = \frac{-8}{6} = -\frac{4}{3} \), 2. \( u_2 = \frac{-22 - 14}{6} = \frac{-36}{6} = -6 \). Thus, the solutions for \( u \) are: \[ -\frac{4}{3}, -6. \]
