Answer:
x = 3
Option 1 is correct.
Step-by-step explanation:
Given: [tex]\dfrac{3}{x-2}+6=\sqrt{x-2}+8[/tex]
First we check restriction on x from equation.
Denominator can't be 0
So, x-2 ≠0
x ≠ 2
Inside the square root value always positive.
Therefore, x-2 ≥ 0
x ≥ 2
Now simplify the equation.
[tex]\dfrac{3}{x-2}+6=\sqrt{x-2}+8[/tex]
[tex]\dfrac{3}{x-2}-2=\sqrt{x-2}[/tex]
squaring both sides
[tex](\dfrac{3}{x-2}-2)^2=(\sqrt{x-2})^2[/tex]
[tex]\dfrac{9}{(x-2)^2}+4-\dfrac{12}{x-2}=x-2[/tex]
[tex]9+4(x-2)^2-12(x-2)=(x-2)^3[/tex]
[tex]9+4x^2+16-16x-12x+24=x^3-8-6x^2+12x[/tex]
[tex]x^3-10x^2+40x-57=0[/tex]
[tex](x-3)(x^2-7x+19)=0[/tex]
x=3 and two are imaginary
Now we take intersection of all two condition and one solution.
x ≠ 2 and x ≥ 2 and x=3
Intersection of all three, x = 3
Hence, The solution of the given equation is 3
