$${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$

Let $$3{x^2} + 4x + 3 = a$$

and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$

Given equation becomes

$$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$

$$ \Rightarrow a(a - kb) - b(a - kb) = 0$$

$$ \Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb$$ or a = b (reject) $$\because$$ a = kb

$$ \Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)$$

$$ \Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0$$ for real roots

D $$\ge$$ 0

$$ \Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0$$

$$ \Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0$$

$$ \Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0$$

$$ \Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0$$

$$ \Rightarrow (k - 1)(2k - 5) \le 0$$



$$\therefore$$ $$k \in \left[ {1,{5 \over 2}} \right]$$

$$\therefore$$ $$k \ne 1$$

$$\therefore$$ $$k \in \left( {1,{5 \over 2}} \right]$$

x² + (20)^1/4x + (5)^1/2 = 0

$$ \Rightarrow $$ x² + $$\sqrt 5$$ = - (20)^1/4x

Squaring both sides, we get

$${\left( {{x^2} + \sqrt 5 } \right)^2} = \sqrt {20} {x^2}$$

$$ \Rightarrow $$ x⁴ = $$-$$5 $$\Rightarrow$$ x⁸ = 25

$$ \Rightarrow $$ $$\alpha$$⁸ + $$\beta$$⁸ = 50

|x|² $$-$$ |x| $$-$$ 12 = 0

$$ \Rightarrow $$ (|x| + 3)(|x| $$-$$ 4) = 0

$$ \Rightarrow $$ |x| = 4

$$\Rightarrow$$ x = $$\pm$$2

$$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$

$$ = \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\} $$

$$ = max\{ {2^{6\sin 3x + 8\cos 3x}}\} $$

and $$\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\} $$

Now range of $$6sin3x + 8cos3x$$

$$ = \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]$$

$$\alpha$$ = 2¹⁰ & $$\beta$$ = 2^$$-$$10

So, $$\alpha$$^1/5 = 2² = 4

$$\Rightarrow$$ $$\beta$$^1/5 = 2^$$-$$2 = 1/4

quadratic 8x² + bx + c = 0

$$ - {b \over 8} = {{17} \over 4}$$ $$ \Rightarrow $$ b = -34

$${c \over 8} = 1$$ $$ \Rightarrow $$ c = 8

$$ \therefore $$ c – b = 8 + 34 = 42