Question

HELP ME, PLEASE. 100 POINTS
find x:
$(x-1)\sqrt{x^{2}-2x+4 } +(x-3)\sqrt{x^{2}-6x+12 }+2x-4=0$

Answer 1

Answer:

x=2

Step-by-step explanation:

Answer 2

Answer:

$\huge \boxed{ \boxed{ \tt x = 2}}$

Step-by-step explanation:

to understand this

you need to know about:

• equation

given:

• $(x-1)\sqrt{x^{2}-2x+4 } +(x-3)\sqrt{x^{2}-6x+12 }+2x-4=0 \atop$

to solve:

• x

let's solve:

1. $\sf rewrite \: (x - 1) \sqrt{ {x }^{2} - 2x + 4} \: as \: (x - 2)( \sqrt{ {x}^{2} - 2x + 4 } ) + (x + 1) ( \sqrt{ {x}^{2} - 2x + 4}) : \\ (x-2)\sqrt{x^{2}-2x+4 } +(x + 1)( \sqrt{ {x}^{2} - 2x + 4)} + (x-3)\sqrt{x^{2}-6x+12 }+2x-4=0$
2. $\sf rewrite (x - 3) \sqrt{ {x}^{2} - 6x + 12 } \: as \: (x - 2) \sqrt{ {x}^{2} - 6x + 12} - : \\ (x-2)\sqrt{x^{2}-2x+4 } +(x + 1)\sqrt{ {x}^{2} - 2x + 4} + (x-2)\sqrt{x^{2}-6x+12 } + (x - 1) \sqrt{ {x}^{2} - 6x + 12} +2x-4=0$
3. $(x - 2) \{\sqrt{ {x}^{2} - 2x + 4 } + \sqrt{ {x}^{2} - 6x + 12} + \frac{4}{\sqrt{ {x}^{2} - 2x + 4 } + \sqrt{ {x}^{2} - 6x + 12} } + 2 \} = 0$
5. $\sf \: divide \: both \: sides \: by \: \{\sqrt{ {x}^{2} - 2x + 4 } + \sqrt{ {x}^{2} - 6x + 12} + \frac{4}{\sqrt{ {x}^{2} - 2x + 4 } + \sqrt{ {x}^{2} - 6x + 12} } + 2 \} : \\ x - 2 = 0$
6. $\therefore \: x = 2$