MaxWong

Substitute (x, y) = (81, 3 sqrt(2)).

$3\sqrt 2 = k \cdot 81^{1/4}\\ k = \sqrt 2$

Then, substitute k = sqrt(2) and x = 4.

$y = \sqrt 2 \cdot 4^{1/4} = 2$

So, you just want to solve x^3 + x^2 + x + 1 = 0?

Use the definition of $\star$ to get $5x + 2x - 5 = 37$. Can you take it from here?

Geometrically, |a - b| denotes the distance between the positions of a and b on the complex plane.

The problem translates to finding the maximum distance between two points on a unit circle, which is attained when a and b are endpoints of a diameter of the unit circle.

Can you take it from here?

Since x≠-1 and x≠2, we can multiply both sides by $x^2 - x - 2$.

$2x + 17 = a\cdot (x + 1) + b\cdot (x - 2)$

That means $(a + b)x + (a - 2b) = 2x + 17$. Then $\begin{cases} a + b = 2\\ a - 2b = 17 \end{cases}$.

Can you take it from here?

You can use the point-slope form and simplify the resulting equation.

Point-slope form:

Let $L$ be a straight line with slope $m$passing through $(x_0, y_0)$. 

Then the equation of $L$ is $y - y_0 = m(x - x_0)$.

Can you take it from here?

Let x = BC and y = AM = MB. By Pythagorean theorem we have $\begin{cases}(2y)^2 + x^2 = 7^2\\y^2 + x^2 = 4^2\end{cases}$.

Subtracting the second equation from the first gives an equation solely in y^2.

Can you take it from here?

By remainder theorem, we get $\begin{cases} f(1) = 5\\ f(2) = -53 \end{cases}$.

Now, using the definition of f(x), we have a system of linear equations in a and b. Can you take it from here?

Divide by ab on both sides of the equation. 

$\dfrac ab + \dfrac ba = 4$

Squaring gives 

$\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} + 2 = 16\\ \dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} =14$

Let x = a/b - b/a. Then x^2 = a^2/b^2 + b^2/a^2 - 2 = 12.

Therefore, $\dfrac ab - \dfrac ba = \pm 2 \sqrt 3$. Since a > b > 0, the negative root is rejected. $\dfrac ab - \dfrac ba = 2\sqrt 3$.

Please try to fill in the missing details.

Previously answered here: https://web2.0calc.com/questions/number-theory_68857