+0

# Need a help please

0
474
6
+262

sabi92  Jun 20, 2015

### Best Answer

#4
+27251
+20

Here are the steps:

.

Alan  Jun 20, 2015
#1
+14536
+10

### Hello sabi92,

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

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

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

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

#### Gruß radix !

radix  Jun 20, 2015
#2
+262
0

Sorry I don't understand how you get this

sabi92  Jun 20, 2015
#3
+94203
+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
#4
+27251
+20
Best Answer

Here are the steps:

.

Alan  Jun 20, 2015
#5
+14536
+5

### Alan, thank you very much, you have the same result like me !

#### Gruß radix !

radix  Jun 20, 2015
#6
+27251
+5

Yes; good to get confirmation of the result.

However, it is probably the steps that are more important than the result for someone wanting to learn how to solve these sorts of equations!

.

Alan  Jun 20, 2015

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