Answer:
3rd Option is correct.
Step-by-step explanation:
Given Equation:
x² - 16x + 12 = 0
First We need to find solution of the given equation.
x² - 16x + 12 = 0
here, a = 1 , b = -16 & c = 12
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-16)\pm\sqrt{(-16)^2-4(12)}}{2}[/tex]
[tex]x=\frac{16\pm\sqrt{256-48}}{2}[/tex]
[tex]x=\frac{16\pm\sqrt{208}}{2}[/tex]
[tex]x=\frac{16\pm4\sqrt{13}}{2}[/tex]
[tex]x=8+2\sqrt{13}\:\:andx=8-2\sqrt{13}[/tex]
Now,
Option 1).
( x - 8 )² = 144
x - 8 = ±√144
x - 8 = ±12
x = 8 + 12 = 20 and x = 8 - 12 = -4
Thus, This is not correct Option.
Option 2).
( x - 4 )² = 4
x - 4 = ±√4
x - 4 = ±2
x = 4 + 2 = 6 and x = 4 - 2 = 2
Thus, This is not correct Option.
Option 3).
( x - 8 )² = 52
x - 8 = ±√52
x - 8 = ±2√13
x = 8 + 2√13 and x = 8 - 2√13
Thus, This is correct Option.
Option 4).
( x - 4 )² = 16
x - 4 = ±√116
x - 4 = ±4
x = 4 + 4 = 8 and x = 4 - 4 = 0
Thus, This is not correct Option.
Therefore, 3rd Option is correct.
