1. ## quadratical equation

hello this is my first post on this forum and i hope i am in the right category,

(2x+3)/(x-1)=2+(5/(x-1))

x²-3x+1-(9/4)-3*(3/2)+1=(x-(3/2))²

2. Originally Posted by beethoven
hello this is my first post on this forum and i hope i am in the right category,

(2x+3)/(x-1)=2+(5/(x-1))

x²-3x+1-(9/4)-3*(3/2)+1=(x-(3/2))²

HI

$\displaystyle \frac{2x+3}{x-1}=\frac{2(x-1)+5}{x-1}$

Multiply both sides by x-1

$\displaystyle 2x+3=2x-2+5$

Continue here .

(2) $\displaystyle x^2-3x+1-\frac{9}{4}-3(\frac{3}{2})+1=(x-\frac{3}{2})^2$

$\displaystyle x^2-3x+1-\frac{9}{4}-\frac{9}{2}+1=x^2-3x+\frac{9}{4}$

Multiply both sides by 4

$\displaystyle 4x^2-12x+4-9-18+4=4x^2-12x+9$

$\displaystyle -12x-23=-12x$

Are you sure this is the correct equation , or else any value of real x would satisfy .

3. yes the equation is false,herse the rigth one:

$\displaystyle x^2-3x+1-(-3/2)=(x-(3/2))^2$

4. Originally Posted by beethoven
hello this is my first post on this forum and i hope i am in the right category,

(2x+3)/(x-1)=2+(5/(x-1))

...
$\displaystyle \dfrac{2x+3}{x-1}=2+\dfrac5{x-1}$
Multiply both sides by (x - 1):

$\displaystyle 2x+3 = 2(x-1) + 5~\implies~ 0=0$

That means: This equation is true for all $\displaystyle x\in \mathbb{R} \setminus \{1\}$

5. Originally Posted by beethoven
yes the equation is false,herse the rigth one:

$\displaystyle x^2-3x+1-(-3/2)=(x-(3/2))^2$
$\displaystyle x^2-3x+\frac{5}{2} = x^2-3x+\frac{9}{4}$

Multiply by 4

$\displaystyle (4x^2-12x+10)-(4x^2-12x+9) = 0$

$\displaystyle 10 - 9 \neq 0$ which means there are no solutions

You sure that's correct?