Question

HELP ME, PLEASE.

find x:

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

Answer 1

x=2

Explanation:

Answer 2

Explanation:

x=2

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`

`\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`

1. 
   `(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`

1. 
   `\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`
2. 
   `\therefore \: x = 2`