The quadratic equation whose roots are reciprocal of 2x² + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)² + 5(1/x) + 3 = 0
=> 3x² + 5x + 2 = 0

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x² + 5x = 0
multiplying both sides by -1/10x
=> x² + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

a³ = 1331 => a = 11
b² = 121 => b = ± 11
a ≥ b

I. a² - 13a + 42 = 0
=>(a - 6)(a - 7) = 0 => a = 6, 7
II. b² - 15b + 56 = 0
=>(b - 7)(b - 8) = 0 => b = 7, 8
=>a ≤ b

x² + 7x - 6x + 42 = 0
x(x + 7) - 6(x + 7) = 0
(x + 7)(x - 6) = 0 => x = -7, 6

2x² + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2

3x² - 9x + 2x - 6 = 0
3x(x - 3) + 2(x - 3) = 0
(x - 3)(3x + 2) = 0 => x = 3, -2/3

I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b

I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b

I. (a - 4)(a + 2) = 0
=> a = 4, -2
II. b² = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.