Solve a^2-12a+16

$$a ^ { 2 } - 12 a + 16$$

Factor

$$\left(a-\left(6-2\sqrt{5}\right)\right)\left(a-\left(2\sqrt{5}+6\right)\right)$$

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.

$$a^{2}-12a+16=0$$

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.

$$a=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 16}}{2}$$

Square -12.

$$a=\frac{-\left(-12\right)±\sqrt{144-4\times 16}}{2}$$

Multiply -4 times 16.

$$a=\frac{-\left(-12\right)±\sqrt{144-64}}{2}$$

Add 144 to -64.

$$a=\frac{-\left(-12\right)±\sqrt{80}}{2}$$

Take the square root of 80.

$$a=\frac{-\left(-12\right)±4\sqrt{5}}{2}$$

The opposite of -12 is 12.

$$a=\frac{12±4\sqrt{5}}{2}$$

Now solve the equation a=\frac{12±4\sqrt{5}}{2} when ± is plus. Add 12 to 4\sqrt{5}.

$$a=\frac{4\sqrt{5}+12}{2}$$

Divide 12+4\sqrt{5} by 2.

$$a=2\sqrt{5}+6$$

Now solve the equation a=\frac{12±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 12.

$$a=\frac{12-4\sqrt{5}}{2}$$

Divide 12-4\sqrt{5} by 2.

$$a=6-2\sqrt{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6+2\sqrt{5} for x_{1} and 6-2\sqrt{5} for x_{2}.

$$a^{2}-12a+16=\left(a-\left(2\sqrt{5}+6\right)\right)\left(a-\left(6-2\sqrt{5}\right)\right)$$

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.

$$x ^ 2 -12x +16 = 0$$

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

$$r + s = 12 rs = 16$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 6. 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>

$$r = 6 - u s = 6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 16

$$(6 - u) (6 + u) = 16$$

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

$$36 - u^2 = 16$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 16-36 = -20$$

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

$$u^2 = 20 u = \pm\sqrt{20} = \pm \sqrt{20}$$

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

$$r =6 - \sqrt{20} = 1.528 s = 6 + \sqrt{20} = 10.472$$

Evaluate

$$a^{2}-12a+16$$