? x² + x ? 2 = 0
=>x²-x+2 = 0
D=(-1)²-4*1*2 = 1-8 = -7 ≤ 0.
Since D ≤ 0, imaginary roots are re.
=>x = \(\frac{1±\sqrt{D}}{2.1} = \frac{1±i\sqrt{7}}{2}\).

Since, p and q are roots of equation x² + px + q =0
So, p + q = -p and pq = q
So, pq = q
And, q = 0 or p = 1
If, q = 0 n, p = 0 and if p = 1 n q = -2.

Since roots of given equation are of opposite sing,
So, products f roots < 0
p(p – 1) / 3 < 0
p(p – 1) < 0
0 < p < 1
For real and distinct roots ½ – √21/ 6 < p < ½ + √21/6

Here, f(x) = x⁴ + 4x³ + 6ax² + 6bx + c so, let its roots be, α, β, γ, δ and
g(x) = x³ + 3x² + 9x + 3 so, let its roots be, α, β, γ
So, from here we can conclude,
α + β + γ + δ = -4 and α + β + γ = -3
Thus, δ = -1
This means (x + 1)(x³ + 3x² + 9x + 3)
On solving this equation in simpler form we get,
x⁴ + 4x³ + 12x² + 12x + 3
=> 6a = 12 => a = 2
=> 6b = 12 => b = 2
=> c = 3
=> a + b + c = 7

On multiplying reciprocal of complex number (1/z) to complex number z, we get multiplying inverse one i.e. z*1=z.

Since complex number x + y i is represented by (x, y) on argand plane, distance of point (x, y) from origin is \(\sqrt{x^2+y^2}\).

We know, i² = -1.
-7 = 7(i²)
Square root of i² is ±i so, square root of -7 are \(\sqrt{7}\)i and –\(\sqrt{7}\)i.

Subtracting equation x² + ax + b = 0 to x² + bx + a = 0 by solving equation simultaneously, we get,
(a ? b)x + (b ? a) = 0
So, (a ? b)x = (a – b)
Therefore, x = 1
Now, putting value of x = 3 in any one of equation, we get,
1 + a + b = 0
Therefore, a + b = -1.

2x²+?2x+2 = 0
=>D=(\(\sqrt{2}\))² – 4.2.2 = 2-16 = -14.
Since D ≤ 0, imaginary roots are re.
=>x = \(\frac{-\sqrt{2}±\sqrt{D}}{2.2} = \frac{-\sqrt{2}±i\sqrt{14}}{4} = \frac{-1±i\sqrt{7}}{2\sqrt{2}}\).

For, imaginary roots, D > 0
Where, D = b² ? 4ac
In equation, (a + 1)x² + 2(a+1)x + (a ? 2) = 0
D = [2(a+1)]² – 4 (a + 1)(a ? 2)
= 4a² + 4 + 8a ? 4{a² ? 2a + a ? 2}
= 4a² + 4 + 8a ? 4a² + 4 a + 8 < 0
=> 12a + 12 < 0
=> 12a < -12
=> a < -1
Therefore, a € (-∞, -1)