Algebra

Let Luis's age be x. Tom's age is x + 2. Product is x(x + 2) = 150. x^2 + 2x = 150, so x^2 + 2x - 150 = 0.

Vertex form: f(x) = a(x-h)^2 + k. Vertex is (1, 3), so f(x) = a(x-1)^2 + 3. Given f(0) = 1, so a(0-1)^2 + 3 = 1, a + 3 = 1, a = -2. f(x) = -2(x-1)^2 + 3. f(10) = -2(10-1)^2 + 3 = -2(81) + 3 = -162 + 3 = -159.

Let 2^(y-2) = x. Then 2^(3-y) = 8/2^y = 8/(4 * 2^(y-2)) = 2/x. The equation becomes 2/x + x - 3 = 0 => x^2 - 3x + 2 = 0. Roots are x=1, x=2. If 2^(y-2) = 1, y=2. If 2^(y-2) = 2, y=3.

Case 1: x^2 - x - 6 = x + 2 => x^2 - 2x - 8 = 0 => (x-4)(x+2)=0 => x=4, -2. Check: |16-4-6|=6, 4+2=6 (Valid); |4+2-6|=0, -2+2=0 (Valid). Case 2: -(x^2 - x - 6) = x + 2 => -x^2 + x + 6 = x + 2 => x^2 = 4 => x=2, -2. Check: |4-2-6|=4, 2+2=4 (Valid). Roots are 4, -2, 2. Total 3 roots.

A quadratic equation has highest power 2 after simplifying. In option D, the cubic terms cancel, leaving x^2 + 2x + 3 = 0, which is quadratic.

Testing for common roots using the resultant or subtraction: (x^3 + 3x^2 + 4x + 5) - (x^3 + 2x^2 + 7x + 3) = x^2 - 3x + 2 = 0. Roots are x=1, x=2. Plugging these into the original equations shows neither is a root.

x^2 - 5x + (2k-1) = 0. Roots r1, r2. r1+r2 = 5, r1-r2 = 3. 2r1 = 8, r1 = 4. r2 = 1. Product r1*r2 = 4 = 2k-1. 2k = 5, k = 2.5.

For 5x^2 - (4p-1)x - (3p+2) = 0, sum of roots = (4p-1)/5 and product = -(3p+2)/5. Setting them equal: 4p - 1 = -3p - 2. 7p = -1. p = -1/7. The question asks for 1/p, which is -7.

Let roots be a and a^2. Product of roots = a * a^2 = a^3 = 8, so a = 2. Sum of roots = a + a^2 = p. p = 2 + 2^2 = 6.

Roots a, b satisfy a+b = 7 and ab = P. Given a-b = 3. (a-b)^2 = (a+b)^2 - 4ab. 3^2 = 7^2 - 4P. 9 = 49 - 4P. 4P = 40, so P = 10.

Let alpha be the common root. alpha^2 - k*alpha - 21 = 0 and alpha^2 - 3k*alpha + 35 = 0. Subtracting: 2k*alpha - 56 = 0, so k*alpha = 28, alpha = 28/k. Substitute into first: (28/k)^2 - k(28/k) - 21 = 0. 784/k^2 - 28 - 21 = 0. 784/k^2 = 49. k^2 = 784/49 = 16. k = 4 or -4.

Interpret the equation as 3^x - 3^(x-1) = 486. Factoring gives 3^(x-1)(3 - 1) = 486, so 3^(x-1) = 243 = 3^5. Therefore x = 6.

Discriminant D = (2a+b)^2 - 4(2a)(b) = 4a^2 + 4ab + b^2 - 8ab = 4a^2 - 4ab + b^2 = (2a-b)^2. Since a and b are integers, the discriminant is a perfect square. Therefore, the roots are rational.