BRIEF ANSWERS FOR TEST 2

• To solve (1+2i)z+(3-4i)=1, subtract (3-4i) from both sides: (1+2i)z=(-2+4i). Now divide through by (1+2i): z = (-2+4i)/(1+2i). Multiply top and bottom of this fraction by (1-2i): z = (6+8i)/(5) = (6/5) + (8/5)i.
• Rewriting each negative square root, square_root(-a), as square_root(a)*i, and then using i^2 = -1, the expression simplifies to 2*square_root(6)*i.
• (-1+i)^2 + 4(-1+i) + 4 = (1-2*i+(-1)) + (-4+4*i) + 4 = 2*i.
• Since z^3-5z^2-4z factors as z(z^2-5z-4), the roots of z^3-5z^2-4z=0 are z = 0 and the roots of z^2-5z-4=0. The latter can be solved with the quadratic formula: z = (5 +/- square_root(41))/2. Thus the three roots of z^3-5z^2-4z=0 are z = 0, (5 +/- square_root(41))/2.
• To find the real solutions of square_root(x+2) = square_root(x) + 1, first square both sides: x+2 = x + 2*square_root(x) + 1. Simplify this to 1 = 2*square_root(x). Square both sides to obtain 1 = 4*x, or x = 1/4. This should be checked in the original equation. If you do, you see that it works, so x = 1/4 is the solution.
• We need to find the solution to |x-1| less-or-equal-to x+2. CASE 1: If x-1 is nonnegative, then this reduces to x-1 less-or-equal-to x+2, or -1 less-or-equal-to 2. This is valid for ALL x in CASE 1: that is, all x greater-or-equal-to 1. Thus, the solution to CASE 1 is: all x greater-or-equal-to 1. Now CASE 2: If x-1 is negative, then the inequality reduces to -(x-1) less-or-equal-to x+2. This simplifies to -1/2 less-or-equal-to x. Since in CASE 2 x is less that 1, the solution in CASE 2 is all x greater-or-equal-to -1/2 and less than 1. Putting the two cases together, we get that the solution set is all real x greater-or-equal-to -1/2. We can express this as [-1/2,+infinity).
• The `critical' points of (w^2-w-6)/(w-6) = (w+2)(w-3)/(w-6) are w = -2, 3 (zeros) and w = 6 (undefined). The sign graph has the form: -----0++++++++0----|++++++ where the changes take place at -2, 3 and 6. Therefore, the expression is greater-or-equal-to zero when w is from [-2,3] union (6,+infinity).
• We want x+y = 12, xy greater-or-equal-to 10, and x less-or-equal-to y. The first two statements combine to x*(12-x) greater-or-equal-to 10, which can be rewritten as 0 greater-or-equal-to x^2-12x+10. Using a sign graph we can solve to find that x is between 6-square_root(26) (about .901) and 6+square_root(26) (about 11.099). But since x is the smaller of two numbers which add up to 12, we must have x less-or-equal-to 6. Thus x must be in [6-square_root(26),6].