Alan

To check if x = 12 is correct, substitute this value back into the original equation and see if the left-hand side equals the right hand side.

As follows:

The following satisfies the conditions

             ABA

          + 411

   =      1215

we must have A + 1 = 5, so A = 4

Hence

          4B4

     +   411

  =    1215

or 4*6² + B*6 + 4 + 4*6² +  1*6 + 1 = 1*6³ + 2*6² + 1*6 + 5 

I'll leave you to calculate B from this.

Let the roots be r and 2r:

(x-r)*(x-2r) = x^2 + 6x + k

x^2 - 3rx + 2r^2 =  x^2 + 6x + k

Compare coefficients

-3r = 6,  hence r = -2

k = 2r^2  Hence k = 8

I get the following:

Factor the cubic polynomial to get \((2x+5)(2x-3)^2\)

This can be written as: \(4(2x+5)(x-\frac{3}{2})^2\)

So r = 3/2

Use:  f = c/lambda   where f is frequency, c is speed of light, lambda is wavelength

        E = hf   where E is energy, h is Planck's constant, f is frequency

Draw a horizontal line through point (2,1) until it is directly below the circle centre. Draw a vertical line from there to the circle centre. You now have a right angled, isosceles triangle with the radius as the hypotenuse (remembering that the hypotenuse here is at 45degrees ). Thus the two legs of this right angled triangle must be of length \(\frac{r}{\sqrt 2}\) by Pythagoras. Add this to the height (1) of the point (2,1).

As follows: