Solutions to the Above Problems

1. 
   Let x = speed of airplane in still air, y = speed of wind and D the distance between A and B. Find the ratio x / y
   Against the wind: D = 8(x - y), with the wind: D = 7(x + y)
   8x - 8y = 7x + 7y, hence x / y = 15
2. 
   Rewrite equations of circles in standard form. Hence equation x² + y² -2x + 4y + 1 = 0 may be written as
   (x - 1)² + (y + 2) ² = 4 = 2²
   and equation x² + y² -2x + 4y - 11 = 0 as
   (x - 1)² + (y + 2) ² = 16 = 4²
   Knowing the radii, the area of the ring is π (4)² - π (2)² = 12 π
3. 
   The given equation is a quadratic equation and has no solutions if it discriminant D is less than zero.
   D = (-m)² - 4(2)(m) = m² - 8 m
   We nos solve the inequality m² - 8 m < 0
   The solution set of the above inequality is: (0 , 8)
   Any value of m in the interval (0 , 8) makes the discriminant D negative and therefore the equation has no real solutions.
4. 
   Write equation in N as follows
   N + 1/N = 78/15
   Multiply all terms by N, obtain a quadratic equation and solve to obtain N = 5.
5. 
   4^m / 125 = 5ⁿ / 64
   
   Cross multiply: 64 4^m = 125 5ⁿ
   Note that 64 = 4³ and 125 = 5³
   The above equation may be written as: 4^{m + 3} = 5^{n + 3}
   The only values of the exponents that make the two exponential expressions equal are: m + 3 = 0 and n + 3 = 0, which gives m = - 3 and n = - 3.
6. 
   3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001} = 3(3^{n + 4001}) = 3^{n + 4002}
7. 
   P(x² + 1) = - 2 x⁴ + 5 x² + 6
   Let t = x² + 1 which also gives x² = t - 1
   Substitute x² by t - 1 in P to obtain: P(t) = - 2 (t - 1)² + 5 (t - 1) + 6 = -2 t ² + 9t - 1
   Now let t = - x² + 3 and substitute in P(t) above to obtain
   P(- x² + 3) = -2 (- x² + 3) ² + 9 (- x² + 3) - 1 = -2 x ⁴ + 3 x ² + 8
8. 
   Solve x + y = r for y: y = r - x
   Substitute in the equation of the circle:
   x² + (r - x)² = 4
   Expand: 2 x² -2 r x + r ² - 4 = 0
   If we solve the above quadratic equation (in x) we will obtain the x coordinates of the points of intersection of the line and the circle. The 2 points of intersection "become one" and therefore the line and the circle become tangent if the discriminant D of the quadratic equation is zero. Hence
   D = (-2r)² - 4(2)(r² - 4) = 4(8 - r²) = 0
   Solve for r to obtain: r = 2 √2 and r = - 2√2