Majid

\(x²+[ (1+\frac{6\sqrt2}{5})x-6]² = 36\)

\(x²+ (1+\frac{6\sqrt2}{5})²x²-12x(1+\frac{6\sqrt2}{5})+36 = 36\)

\(x[(1+ (1+\frac{6\sqrt2}{5})²)x-12(1+\frac{6\sqrt2}{5}) ]= 0\)

so: x=0 and \((1+ (1+\frac{6\sqrt2}{5})²)x-12(1+\frac{6\sqrt2}{5}) = 0\)

\( x =\frac{12(1+\frac{6\sqrt2}{5})}{(1+ (1+\frac{6\sqrt2}{5})²)}\)

\( x =\frac{12(1+\frac{6\sqrt2}{5})}{1+ 1+\frac{72}{25}+\frac{12\sqrt2}{5}}\)

\( x ={\frac{300+360\sqrt2}{122+60\sqrt2}}\)

\(x=3.91\)

Omi67 made a good job, it was the answer :)

9-a 

look the polynomial 2x²+2x-4 has as roots 1 and -2 (if we replace x by 1 or -2 in the polynomial we'll get 0)

For that, we can be factorized, in an other word, the polynomial can be divided by (x-1) or (x+2)

2x²+2x-4=(x-1)(2x+4)

2x²+2x-4=(x+2)(2x-2)

\(\frac{1}{2}\frac{1}{(x+2)} + \frac{1}{x-1} - \frac{x+5}{2x²+2x-4} = \frac{1}{2}\frac{2x-2}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{1}{2}\frac{2(x-1)}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2}{2}\frac{x-1}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{x-1}{2x²+2x-4} + \frac{2x+4}{2x²+2x-4} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2x-2}{2x²+2x-4} = \frac{2(x-1)}{2x²+2x-4} = \frac{2(x-1)}{(x-1)(2x+4) }= \frac{2}{2(x+2) }\)

\(=\frac{1}{(x+2) }\)

\(16x-12x-15=2x+9\)

\(4x-15=2x+9\)

\(2x=24\)

\(x=12\)

Yes, go back to your first post, I also answered your questions.

A- as you can see here, your function f is a referential equation: f(x)=x²

You can notice from the graph, let's call it the graph of the function g, is got by a translation of all the points of the graph of the  referential function f(x) = x² by a vector v(-2,2)

So, according to a theorem, the graph of a function \(x \rightarrow f(x+a)+b \)is got by a translation of the graph of \(x\rightarrow f(x)\)by using the vector v(-a,b)

As a result our function is \(g(x)=(x+2)²+2\)

g(-1) graphically = 3

g(-1) algebrically: g(-1) = (-1+2)²+2 = 1+2 = 3

Let m1 = m2 = m ( the mass of the astronaut )

the force F is equal to (G*m1*m2)/d² where d is the distance between the two astronauts

\(f=\frac{G{m}_{1}{m}_{2}}{d²}\)

\(f=\frac{Gm²}{d²}\)

\(f=\frac{6.67*{10}^{-11}*{10}^{4}}{4} = 1.6675*{10}^{-7} N\)

that means there is an attracted force between any two material objects.

This force is proportionally inversed with the square of the distance separating the two objects

according to the result, this force isn't noticibale (1.6675*10^7 is practically 0), that's why we can't feel it

\(0.25x+3=0.1x+4.5 \)

\(25x+300=10x+450\)

\(15x=150\)

\(x=10\)

thanks Guest!

I actually have the solutions, but I expect to find better ones here because I noticed many times that some solving ways are better than some others :)