Problems

An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.

Find the area between two concentric circles defined by
x^{2} + y^{2} 2x + 4y + 1 = 0
x^{2} + y^{2} 2x + 4y  11 = 0

Find all values of parameter m (a real number) so that the equation 2x^{2}  m x + m = 0 has no real solutions.

The sum an integer N and its reciprocal is equal to 78/15. What is the value of N?

m and n are integers so that 4^{m} / 125 = 5^{n} / 64. Find values for m and n.

Simplify: 3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001}

P is a polynomial such that P(x^{2} + 1) =  2 x^{4} + 5 x^{2} + 6. Find P( x^{2} + 3)

For what values of r would the line x + y = r be tangent to the circle x^{2} + y^{2} = 4?
Solutions to the Above Problems
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

Rewrite equations of circles in standard form. Hence equation x^{2} + y^{2} 2x + 4y + 1 = 0 may be written as
(x  1)^{2} + (y + 2) ^{2} = 4 = 2^{2}
and equation x^{2} + y^{2} 2x + 4y  11 = 0 as
(x  1)^{2} + (y + 2) ^{2} = 16 = 4^{2}
Knowing the radii, the area of the ring is π (4)^{2}  π (2)^{2} = 12 π

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) = m^{2}  8 m
We nos solve the inequality m^{2}  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.
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.

4^{m} / 125 = 5^{n} / 64
Cross multiply: 64 4^{m} = 125 5^{n}
Note that 64 = 4^{3} and 125 = 5^{3}
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.

3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001} = 3(3^{n + 4001}) = 3^{n + 4002}

P(x^{2} + 1) =  2 x^{4} + 5 x^{2} + 6
Let t = x^{2} + 1 which also gives x^{2} = t  1
Substitute x^{2} 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 =  x^{2} + 3 and substitute in P(t) above to obtain
P( x^{2} + 3) = 2 ( x^{2} + 3) ^{2} + 9 ( x^{2} + 3)  1 = 2 x ^{4} + 3 x ^{2} + 8

Solve x + y = r for y: y = r  x
Substitute in the equation of the circle:
x^{2} + (r  x)^{2} = 4
Expand: 2 x^{2} 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)(r^{2}  4) = 4(8  r^{2}) = 0
Solve for r to obtain: r = 2 √2 and r =  2√2
