Question

X^2 - 6 x + 9 = 4 root(x^2 - 6 x + 6)​

Answer 1

Answer

X1= 3-2 root3 x2= 1 x3=5 x4= 3 +2 root 3
First swap the sides then simplify the equation
Collect like terms
Move all the expression to the left to equal 0
Then collect like terms again
Reorder the terms from ^2 to ^4
Factor the expression
Change the signs
Separate into possible cases
Then sold the equation to equal 0
Then check solutions by subbing them in as x
Then you should have 4 perfect answers

Answer 2

Answer:

Hello,

Step-by-step explanation:

$x^2-6x+9=4\sqrt{x^2-6x+6} \\\\(x-3)^2=4\sqrt{x^2-6x+9-3} \\\\(x-3)^2=4\sqrt{(x-3)^2-3} \\\\Let's\ say\ y=x-3\\\\y^2=4\sqrt{y^2-3} \ squaring\\\\y^4=16(y^2-3)\\y^4-16y^2+48=0\\\Delta=16^2-4*48=64=8^2\\y^2=4\ or\ y^2=12\\\\(y-2)(y+2)=0\ or\ (y+2\sqrt{3} )(y-2\sqrt{3} )=0 \\\\y=2,x=5\ or\\\ y=-2, x=-1\ or\\\ y=-2\sqrt{3},x=3-2\sqrt{3}\ \\or\ y=2\sqrt{3},x=3+2\sqrt{3}$

Verifying:

$1) if\ x=-1\ then\\a)\ x^2-6x+9=1+6+9=16\\b)\ 4\sqrt{x^2-6x+6} =4\sqrt{1+6+6} =4\sqrt{13}:\ False\\\\2)if\ x=5\\a)\ x^2-6x+9=25-30+9=4\\b)\ 4\sqrt{x^2-6x+6} =4\sqrt{25-30+6} =4\sqrt{1}=4:\ True$

$3) if\ x=3-2\sqrt{3}\ x^2=21-12\sqrt{3} \\a)\ x^2-6x+9=21-12\sqrt{3}-18+12\sqrt{3}+9=12\\b)\ 4\sqrt{x^2-6x+6} =4\sqrt{21-12\sqrt{3}-18+12\sqrt{3}+6} =4\sqrt{9}=12:\ True\\\\\\4) if\ x=3+2\sqrt{3}\ x^2=21+12\sqrt{3} \\a)\ x^2-6x+9=21+12\sqrt{3}-18-12\sqrt{3}+9=12\\b)\ 4\sqrt{x^2-6x+6} =4\sqrt{21+12\sqrt{3}-18-12\sqrt{3}+6} =4\sqrt{9}=12:\ True$

Sol={5,3-2√3,3+2√3}