#1**+10 **### Hello sabi92,

#### it is too much to write, but here is the answer:

#### Gruß radix !

$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} = {\mathtt{3}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{7}} = {\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{27}}$$

$${\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{20}} = {\mathtt{0}}$$ => $${\mathtt{x1}} = -{\mathtt{2}}$$ ; $${\mathtt{x2}} = {\frac{{\mathtt{10}}}{{\mathtt{7}}}}$$

radix
Jun 20, 2015

#3**+10 **

$$\begin{array}{rlll}

\frac{x-2}{2x+2}-\frac{1}{3x^2-6x-9}-\frac{x-30}{18-6x}&=&3\\\\

\frac{x-2}{2(x+1)}-\frac{1}{3(x^2-2x-3)}-\frac{x-30}{6(3-x)}&=&3\\\\

\frac{x-2}{2(x+1)}-\frac{1}{3(x-3)(x+1)}+\frac{x-30}{6(x-3)}&=&3\\\\

\frac{3(x-3)(x-2)}{3(x-3)*2(x+1)}-\frac{2*1}{2*3(x-3)(x+1)}+\frac{(x+1)(x-30)}{6(x+1)(x-3)}&=&3\\\\

3(x-3)(x-2)-2*1+(x+1)(x-30)&=&3*6(x+1)(x-3)\\\\

3(x^2-5x+6)-2+(x^2-29x-30)&=&18(x^2-2x-3)\\\\

3x^2-15x+18-2+x^2-29x-30&=&18x^2-36x-54\\\\

3x^2+x^2-15x-29x+18-2-30&=&18x^2-36x-54\\\\

4x^2-44x-14&=&18x^2-36x-54\\\\

0&=&14x^2+8x-40\\\\

0&=&7x^2+4x-20\\\\

\end{array}$$

$$\\$Now I need 2 numbers that multiply to give 7*-20=-140$\\

$and add to give +4$\\

$Obviously one will be negative and one will be positive,$\\

$The bigger one will be positive$\\

$And the absolute value difference will be 4$\\

140=7*20\\

140=7*2*2*5\\

140=14*10\\

$The numbers that we want are +14 and -10$\\

$replace 4x with 14x-10x$\\

0&=&7x^2+4x-20\\

7x^2+14x-10x-20=0\\

$Now factor in pairs$\\

7x(x+2)-10(x-2)=0\\

(7x-10)(x-2)&=&0\\

7x-10=0\quad or \quad x-2=0\\

7x=10\quad or \quad x=2\\

x=1\frac{3}{7}\quad or \quad x=2\\$$

Ofcourse you could have just used the quadratic equation to get the final answer but if you wanted to factor it than is the best way to go about it. IMHO

Thanks Alan for spotting my error :)

Melody
Jun 20, 2015