Saved Bookmarks
| 1. |
Solve the in equation √(x + 2) > √(8 − x2). |
|
Answer» When a, b ∈ R and a ≥ 0, b ≥ 0 then √a > √b ⇔ a > b ≥ 0 ∴ √(x + 2) > √(8 - x2) ⇔ x + 2 > 8 – x2 ≥ 0 and x > –2, |x| < 2√2 We have (x + 2) > 8 – x2 ⇔ x2 + x – 6 > 0 ⇔ (x + 3) (x – 2) > 0 ⇔ x ∈ (– ∞, – 3) ∪ (2, ∞) We have 8 – x2 ≥ 0 ⇔ x2 ≤ 8 ⇔ |x| < 2√2 ⇔ x ∈ [–2√2 , 2√2] Also x + 2 > 0 ⇔ x > – 2 Hence x + 2 > 8 – x2 ≥ 0 ⇔ x ε ((– ∞, –3) ∪ (2, ∞)) ∩ [–2√2 , 2√2] and x > –2 ⇔ x ∈ (2, 2√2) The solution set is [x ∈ R : –2 < x ≤ 2√2] |
|