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(400d² - 480000)/(d² + 300) < 0 → you know that a square is always ≥ 0 → (d² + 300) > 0
400d² - 480000 < 0
400.(d² - 1200) < 0 → you know that: 400 > 0
d² - 1200 < 0
d² < 1200
- √1200 < d < √1200 → recall: 1200 = 400 * 3 = 20² * 3
- 20√3 < d < 20√3
-20 Sqrt[3]<d<20 Sqrt[3]
_________________________
(400 d^2-480000)/(d^2+300) < 0 ->
(400 d^2-480000) < 0 (d^2+300) ->
(400 d^2-480000) < 0 ->
400 d^2-480000 < 0 ->
400 d ^2 < 480000 ->
d^2 < 480000/400 ->
d^2 < 1200 ->
- Sqrt[1200] < d < Sqrt[1200] ->
- Sqrt[400 * 3] < d < Sqrt[400 * 3] ->
- 20 Sqrt[3] < d < 20 Sqrt[3]
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Answers & Comments
Te dejo un vídeo step por step
https://youtu.be/JwODg9lu-WI
(400d² - 480000)/(d² + 300) < 0 → you know that a square is always ≥ 0 → (d² + 300) > 0
400d² - 480000 < 0
400.(d² - 1200) < 0 → you know that: 400 > 0
d² - 1200 < 0
d² < 1200
- √1200 < d < √1200 → recall: 1200 = 400 * 3 = 20² * 3
- 20√3 < d < 20√3
-20 Sqrt[3]<d<20 Sqrt[3]
_________________________
(400 d^2-480000)/(d^2+300) < 0 ->
(400 d^2-480000) < 0 (d^2+300) ->
(400 d^2-480000) < 0 ->
400 d^2-480000 < 0 ->
400 d ^2 < 480000 ->
d^2 < 480000/400 ->
d^2 < 1200 ->
- Sqrt[1200] < d < Sqrt[1200] ->
- Sqrt[400 * 3] < d < Sqrt[400 * 3] ->
- 20 Sqrt[3] < d < 20 Sqrt[3]