Solve x^2-4x-4

$$x ^ { 2 } - 4 x - 4$$

Factor

$$\left(x-\left(2-2\sqrt{2}\right)\right)\left(x-\left(2\sqrt{2}+2\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-4x-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-4\right)}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+16}}{2}$$

Add 16 to 16.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2}$$

Take the square root of 32.

$$x=\frac{-\left(-4\right)±4\sqrt{2}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{4±4\sqrt{2}}{2} when ± is plus. Add 4 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+4}{2}$$

Divide 4+4\sqrt{2} by 2.

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

Now solve the equation x=\frac{4±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from 4.

$$x=\frac{4-4\sqrt{2}}{2}$$

Divide 4-4\sqrt{2} by 2.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2+2\sqrt{2} for x_{1} and 2-2\sqrt{2} for x_{2}.

$$x^{2}-4x-4=\left(x-\left(2\sqrt{2}+2\right)\right)\left(x-\left(2-2\sqrt{2}\right)\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

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

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 4 rs = -4$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(2 - u) (2 + u) = -4$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$4 - u^2 = -4$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -4-4 = -8$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 8 u = \pm\sqrt{8} = \pm \sqrt{8}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =2 - \sqrt{8} = -0.828 s = 2 + \sqrt{8} = 4.828$$

Evaluate

$$x^{2}-4x-4$$