# Tutor profile: Pamarthi V.

## Questions

### Subject: Geometry

If the circle x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally, then k

For the circles to intersect orthogonally, we must have 2g1g2 + 2f1f2 = c1 + c2 Here, g1 = 1, g2 = 0, f1 = k, f2 = k, c1 = 6, c2 = k Hence, we have the equation 2(1)(0) + 2k.k = 6 + k This gives 2k2 – k – 6 = 0 which yields k = -3/2, 2.

### Subject: Basic Math

Factor: 5x2 – 15x – 20

5x2 – 15x – 20. = 5(x2 – 3x – 4). = 5(x2 – 4x + x – 4). = 5{x(x - 4) +1(x - 4)}. = 5(x-4)(x+1).

### Subject: Algebra

The set of all real numbers x for which x2 - |x + 2| + x > 0 is ?

The condition given in the question is x2 - |x + 2| + x > 0 Two cases are possible: Case 1: When (x+2) ≥ 0. Therefore, x2 - x - 2 + x > 0 Hence, x2 – 2 > 0 So, either x < - √2 or x > √2. Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1) Case 2: When (x+2) < 0 Then x2 + x + 2 + x > 0 So, x2 + 2x + 2 > 0 This gives (x+1)2 + 1 > 0 and this is true for every x Hence, x ≤ -2 or x ∈ (-∞, -2) ...…… (2) From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

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