MaxWong

414 is divisible by 9

=> 414n is divisible by 9

=> 414n + 9 is divisible by 9

414n + 12 = (414n + 9) + 3

414n + 12 is 3 more than a multiple of 9. The required remainder is 3.

Let \(\alpha, \beta\) be the roots.

Then the quadratic is \(x^2 - (\alpha + \beta)x + \alpha \beta\).

Set \(\begin{cases} \alpha = -(\alpha + \beta) - 2\\ \beta = \alpha\beta - 2 \end{cases}\).

Then \(\begin{cases} \alpha = \dfrac{-\beta - 2}2\\ \beta = \alpha \beta - 2 \end{cases}\).

Substituting, we have

\(\beta = \beta\left(\dfrac{-\beta-2}{2}\right) - 2\\ \beta = -\dfrac{\beta^2}2 - \beta - 2\\ \dfrac{\beta^2}2 + 2\beta + 2 = 0\\ \beta^2 + 4\beta + 4 = 0\\ (\beta + 2)^2 = 0\\ \beta = -2\)

Now, \(\alpha = \dfrac{-(-2) - 2}2 = 0\)

The required quadratic is \(x^2 - (0 + (-2))x + 0(-2)\)

Simplifying, the required quadratic is \(x^2 + 2x \)

Let x be the length of CQ. Then DQ = 30 - x.

By power of points,

x(30 - x) = 72

x^2 - 30x + 72 = 0

(x - 15)^2 - 225 + 72 = 0

(x - 15)^2 = 153

\(x = 15\pm \sqrt{153}\)

Minimum length of CQ = \(15 - \sqrt{153}\)

You should know how to find u + v and uv.

Hint: u/v + v/u = (u^2 + v^2)/(uv) = ((u + v)^2 - 2uv)/(uv)

The rest is just substituting the values of u + v and uv into the expression.

We first find the intersection point of x^2 - 12x + y^2 = 28 and y = 8 - x.

\(\begin{cases} x^2 - 12x + y^2 = 28\\ y = 8-x \end{cases}\)

After some work, we get (x, y) = \((7-\sqrt{31}, 1+\sqrt{31})\). We don't need the other intersection point.

We also get that the region is represented by the compound inequality \(\max(0, 8- x)\leq y \leq \sqrt{28 + 12x - x^2}\).

We use the straight line x = 8 to divide the region into 2 parts.

Using integration, the area is 

\(\displaystyle \int^{8}_{7-\sqrt{31}}\left(\sqrt{28 +12x - x^2}-(8-x)\right)\,dx+\int^{14}_8 \sqrt{28 + 12x - x^2}\,dx\)

which would be \(16\pi + 32\arcsin\dfrac{\sqrt{31} - 1}{8} - \sqrt{31} - 1 \approx 63.14\text{ sq.units}\)

The computation is too long so I didn't include it. 

Here's a verification of my answer by Wolfram Alpha:

Notice a pattern on the partial sums:

1 = 1 = 1^2

1 + 3  = 4 = 2^2

1 + 3 + 5 = 9 = 3^2

1 + 3 + 5 + 7 = 16 = 4^2

The sum is equal to n^2, where n is the number of terms. (This can be proven by mathematical induction.)

Now, 144 = 12^2. This means n^2 = 12^2, n = 12.

f(x) = f(2x + 1)

5x - 12 = 5(2x + 1) - 12

5x = 5(2x + 1)

x = 2x + 1

-x = 1

x = -1

Each number is a 3-digit number.

This means a + b + c + d = 410 where a,b,c,d >= 100.

Let a' = a - 100, b' = b - 100, etc.

Then a' + b' + c' + d' = 10, where a,b,c,d >= 0.

Then the required number of ways is \(H^4_{10} =\displaystyle \binom{10 + 4-1}{10} = \binom{13}{10}=286\)

The coefficients are rational => Sum of roots and product of roots are also rational.

=> conjugates of 2 + sqrt 2 and 1 - sqrt 5 must also be roots of the polynomial.

=> The roots are 2 + sqrt 2, 2 - sqrt 2, 1 - sqrt 5, 1 + sqrt 5.

The required monic quartic polynomial is 

\(\quad(x - (2 + \sqrt 2))(x - (2 - \sqrt 2))(x - (1 - \sqrt 5))(x - (1 + \sqrt 5))\\ =(x^2 - 4x + 2)(x^2 - 2x - 4)\\ =x^4 - 6x^3 +6x^2 +12x-8\)

The dependent variable is v because v is dependent on b and h.