Expand (x - 4)^2:
[tex] (x - 4) \cdot (x - 4) = (x \cdot x) + (x \cdot -4) + (-4 \cdot x) + (-4 \cdot -4) [/tex]
[tex] x^2 - 4x - 4x + 16 = \boxed{x^2 - 8x + 16 = 12} [/tex]
Subtract 12 from both sides to get one side to equal 0:
[tex] x^2 - 8x + 4 = 0 [/tex]
Find the values of a, b, and c in this quadratic equation:
[tex] x^2 \ | \ a = 1 [/tex]
[tex] -8x \ | \ b = -8 [/tex]
[tex] 4 \ | \ c = 4 [/tex]
The quadratic formula is expressed as follows:
[tex] \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} [/tex]
Plug in our values into the formula:
[tex] \begin{array}{*{20}c} {x = \frac{{ 8 \pm \sqrt {(-8)^2 - 4(1)(4)} }}{{2(1)}}} \end{array} [/tex]
[tex] \begin{array}{*{20}c} {x = \frac{{ 8 \pm \sqrt {64 - 16} }}{{2}}} \end{array} [/tex]
Simplify the square root:
[tex] \sqrt{64 - 16} = \sqrt{48} [/tex]
Prime factorize the square root:
[tex] \sqrt{48} = \sqrt{4 \cdot 12} = \sqrt{2 \cdot 2 \cdot 3 \cdot 4} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} [/tex]
Take any number that is repeated twice in the square root, and move it outside:
[tex] \sqrt{2 \cdot 2} = 2 [/tex]
[tex] \sqrt{(2 \cdot 2) \cdot (2 \cdot 2) \cdot 3} = 2 \cdot 2 \sqrt{3} = \boxed{4 \sqrt{3}} [/tex]
[tex] \begin{array}{*{20}c} {x = \frac{{ 8 \pm 4 \sqrt{3} }}{{2}}} \end{array} [/tex]
Solve the plus and minus:
[tex] \frac{8 + 4 \sqrt{3}}{2} = \boxed{4 + 2\sqrt{3}} [/tex]
[tex] \frac{8 - 4 \sqrt{3}}{2} = \boxed{4 - 2\sqrt{3}} [/tex]
[tex] \boxed{x = 4 + 2\sqrt{3} \ \& \ 4 - 2\sqrt{3}} [/tex]
The answer is {4 + 2√3, 4 - 2√3}.
