Alan

Looks like the solutions are x = -1, x = 0 and x = 1

\(\text{Let } z=a+bi\text{ and }w=c+di\\\text{Then }|z|=\sqrt{a^2+b^2}\text{, }|w|=\sqrt{c^2+d^2} \text{ and }z\bar w=(a+bi)(c-di)=ac+bd+(bc-ad)i\)

\(\text{Hence }a^2+b^2=5^2,ac+bd=6,c^2+d^2=2^2,bc-ad=8\)

(1)

  \(|z+w|=|a+c+(b+d)i|=\sqrt{a^2+c^2+2ac+b^2+d^2+2bd}=\sqrt{a^2+b^2+c^2+d^2+2(ac+bd)}\\=\sqrt{5^2+2^2+2\times6}=\sqrt{41}\\\text{So }|z+w|^2=41\)

See if you can do the rest using a similar approach.  

You need to consider the order in which they could be born, so the possibilities are:

1st born. 2nd born.  3rd born.

    B.          B.              B

    B.          B.              G

    B.          G.              B

    G.          B.              B

    B.          G.              G

    G.          B.              G

    G.          G.              B

    G.          G.              G

So the probability of exactly 2 girls and 1 boy is 3/8.

The force is inversely proportional to the square of the distance between them, so:

F2/F1 = (r1/r2)².  

Here you have r1 = 5cm, r2 = 1cm, F1 = 1N

Can you take it from here?

First, Outer, Inner, Last:  when multiplying two binomials together.

If a = b = c = d = √2 and e = f = g = h = √3, then (ae)² + (bf)² + (cg)² + (dh)² = 24

Not sure I can prove this a minimum though!

Here are a couple of possibilities:

There are four points.

The slope of the line is (281-17)/(48-3) = 88/15, so for each 15 lattice points to the right of 3 you move up by 88 lattice points. 

Perhaps the following will help:

Try adding a negative number, and note that 0.4 is 2/5.

Alternatively, solve (5+x)/(8+x) = 0.4 for x.

(Too late! Guest has already supplied the answer.)