To get rid of the square root, square both terms.
[tex]x+2 = (x-4)^2[/tex]
[tex]x+2 = x^2-8x+16[/tex]
Move all the terms to the left of equal with changed sign
[tex]x+2 - x^2+8x-16 = 0[/tex]
Sum like terms.
[tex]-x^2 + 9x - 14 = 0[/tex]
Multiply all the terms for -1 (change their sign) to make the x positive
[tex]x^2 - 9x + 14 = 0[/tex]
We have a second grade equation, that we'll solve in this way.
x₁,₂ = (-b±√Δ)/2a
Δ = b²-4ac
You call a (1)x², you call b 9 and you call c 14
x1,2 = (-(-9)±√(-9)² - 4*1*14)/2*1
x1,2 = (9±√81-56)/2
x1,2 = (9±√25)/2
x1,2 = (9±5)/2
x1 = (9+5)/2 = 14/2 = 7
x2 = (9-5)/2 = 4/2 = 2
Check what x is true for √x+2 = x-4
√7+2 = 7-4
√9 = 3
3 = 3 (true)
√2+2 = 2-4
√4 = -2
2 = -2 (not true)
Therefore the only solution is x = 7
Answer C
