3) I'm gonna assume it means a double root, so the equation is (x - 2-3 sq 5)² = 0 -> x² - (4 (plus) 6sq5)x (plus) shit squared

cool dude, im trying to put his in terms my teacher will accept He wants things his way... :(

4) When it travels upstream the current is substracted from the speed, when it travels downstream it's added, so

65/(x-5) pl 55/(x pl 5) = 15

Long equation, yields x² -8x -425/15 -> 10.65ish km/h

Ummm on the second one I'm pretty sure "p" is (positive or negative) 5/2 not 25/4 as zexen said

The third one could be X^2 - 4X - 41

1)

If the initial # of members = x

720/x = y ----> From this, 720 = xy

720/(x-2) = [y (plus) 4] ----> From this  720 = (x-2)[y(plus)4] =  xy (plus) 4x - 2y - 8

Since they're both equal to 720,

xy = xy (plus) 4x - 2y - 8

0 = 4x - 2y - 8

2y = 4x - 8

y = 2x - 4 Substitute this in 720 = xy and we get,

720 = x(2x-4)

720 = 2x² - 4x

Divide everything by 2,

360 = x² - 2x

x² - 2x - 360 = 0

(x plus 18)(x - 20) = 0

Therefore, x = -18 or x=20

Since -18 cannot be taken, the # of members is 20 before, 18 after.

2)

Don't know if you've done this at school but to check the roots of a quadratic equation, you can use this thing called the discriminant (Think thats what its called)

Discriminant = (coefficient of the "x" term)² - 4(coefficient of the x2 term)(coefficient of the constant term)

If its equal to 0, then the equation has real, equal roots.

If its more than 0, then there are 2 real, different roots.

If its less than 0, the roots are not real... I think

Anyway, we have to consider the discriminant to be equal to 0.

Coefficient of the,

x term - -5

x² term - p

constant term - p

Therefore, for the equation to have real roots,

(-5)2 - 4(p)(p) = 0

25 - 4p² = 0

25/4 = p²

p = 5/2 or p = -5/2

3)

I don't understand what you're saying....

I'm gonna assume one root is 2 plus 3 sqr.5 and the other is 2 - 3 sqr.5

So the equation is -

(x-[2 plus 3sqr.5])(x-[2 - 3sqr.5]) = 0

x² -[2 - 3sqr.5]x - [2 plus 3sqr.5]x plus [2 plus 3sqr.5][2 - 3sqr.5] = 0

x² - 2x plus 3sqr.5x - 2x - 3sqr.5x plus (22 - 3^2(sqr.5)^2) = 0 (^2 = squared btw)

x²  - 4x -4 plus 4 - 45 = 0

^x2 - 4x - 41 = 0

God I hope that correct....

4)

If the boat is travelling at x km/h,

[65/(x - 5)] plus [55/(x plus 5)] = 15

[65(x plus 5) plus 55(x - 5)] that whole thing divided by (x plus)(x-5) = 15

Divide both sides by 5 to get

[13(x plus 5) plus 11(x - 5)] that whole thing divided by (x plus)(x-5) = 3

[13x plus 65 plus 11x - 55] divided by x² - 25 = 3

24x plus 10 = 3x² - 75

3x² - 24x - 85 = 0

x² - 8x - 85/3 = 0

x² - 8x plus 16 - 133/3

(x - 4)² = 133/3

x - 4 = sqr.(133/3)

x - 4 = 6.658

x = 10.658

There are bound to be some mistakes cause typing this shit in and no plus symbols is pain in the ass

Great...superscript doesn't work either...

p2, x2 and all that shit is squared btw.

Use ^, it's understood to be super

How long did it take you to write that? O_O

All are correct and well explained, congrats

10 minutes ?

No solution in real numbers, but in hyperreal it would be "minus infinite"