Question

Determine the solution set of (x - 4)^2 = 12.

Answer 1

Expand (x - 4)^2:

$(x - 4) \cdot (x - 4) = (x \cdot x) + (x \cdot -4) + (-4 \cdot x) + (-4 \cdot -4)$

$x^2 - 4x - 4x + 16 = \boxed{x^2 - 8x + 16 = 12}$

Subtract 12 from both sides to get one side to equal 0:

$x^2 - 8x + 4 = 0$

Find the values of a, b, and c in this quadratic equation:

$x^2 \ | \ a = 1$

$-8x \ | \ b = -8$

$4 \ | \ c = 4$

The quadratic formula is expressed as follows:

$\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}$

Plug in our values into the formula:

$\begin{array}{*{20}c} {x = \frac{{ 8 \pm \sqrt {(-8)^2 - 4(1)(4)} }}{{2(1)}}} \end{array}$

$\begin{array}{*{20}c} {x = \frac{{ 8 \pm \sqrt {64 - 16} }}{{2}}} \end{array}$

Simplify the square root:

$\sqrt{64 - 16} = \sqrt{48}$

Prime factorize the square root:

$\sqrt{48} = \sqrt{4 \cdot 12} = \sqrt{2 \cdot 2 \cdot 3 \cdot 4} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3}$

Take any number that is repeated twice in the square root, and move it outside:

$\sqrt{2 \cdot 2} = 2$

$\sqrt{(2 \cdot 2) \cdot (2 \cdot 2) \cdot 3} = 2 \cdot 2 \sqrt{3} = \boxed{4 \sqrt{3}}$

$\begin{array}{*{20}c} {x = \frac{{ 8 \pm 4 \sqrt{3} }}{{2}}} \end{array}$

Solve the plus and minus:

$\frac{8 + 4 \sqrt{3}}{2} = \boxed{4 + 2\sqrt{3}}$

$\frac{8 - 4 \sqrt{3}}{2} = \boxed{4 - 2\sqrt{3}}$

$\boxed{x = 4 + 2\sqrt{3} \ \& \ 4 - 2\sqrt{3}}$

The answer is {4 + 2√3, 4 - 2√3}.

Answer 2

Answer:

x = 4 + 2 sqrt(3) or x = 4 - 2 sqrt(3) thus {4 + 2√3, 4 - 2√3} is your answer!

Step-by-step explanation:

Solve for x over the real numbers:

(x - 4)^2 = 12

Take the square root of both sides:

x - 4 = 2 sqrt(3) or x - 4 = -2 sqrt(3)

Add 4 to both sides:

x = 4 + 2 sqrt(3) or x - 4 = -2 sqrt(3)

Add 4 to both sides:

Answer: x = 4 + 2 sqrt(3) or x = 4 - 2 sqrt(3)