Question

helpppppp , quick answer with work pleaseeee​

Answer 1

Answer:

Step-by-step explanation:

$\sqrt{x-2}+8=x\\\\\sqrt{x-2}=x-8$

Square both sides,

$x-2=(x-8)^{2}\\\\x-2=x^{2}-2*x*8+8^{2}\\\\x-2=x^{2}-16x+64\\\\x^{2}-16x+64=x-2\\\\x^{2}-16x+64-x+2=0\\\\x^{2}-17x+66=0$

Sum = - 17

Product = 66

Factors = -6 , -11

x² - 6x -11x + (-6)*(-11) = 0

x(x - 6) -11(x - 6) = 0

(x-6) (x - 11) = 0

x -6 = 0   ; x - 11 = 0

x = 6   ; x =11

Here,  x = 6 is a extraneous solution