Solve y^2-6y+6

$$y ^ { 2 } - 6 y + 6$$

Factor

$$\left(y-\left(3-\sqrt{3}\right)\right)\left(y-\left(\sqrt{3}+3\right)\right)$$

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.

$$y^{2}-6y+6=0$$

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.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 6}}{2}$$

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 6}}{2}$$

Multiply -4 times 6.

$$y=\frac{-\left(-6\right)±\sqrt{36-24}}{2}$$

Add 36 to -24.

$$y=\frac{-\left(-6\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$y=\frac{-\left(-6\right)±2\sqrt{3}}{2}$$

The opposite of -6 is 6.

$$y=\frac{6±2\sqrt{3}}{2}$$

Now solve the equation y=\frac{6±2\sqrt{3}}{2} when ± is plus. Add 6 to 2\sqrt{3}.

$$y=\frac{2\sqrt{3}+6}{2}$$

Divide 6+2\sqrt{3} by 2.

$$y=\sqrt{3}+3$$

Now solve the equation y=\frac{6±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 6.

$$y=\frac{6-2\sqrt{3}}{2}$$

Divide 6-2\sqrt{3} by 2.

$$y=3-\sqrt{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3+\sqrt{3} for x_{1} and 3-\sqrt{3} for x_{2}.

$$y^{2}-6y+6=\left(y-\left(\sqrt{3}+3\right)\right)\left(y-\left(3-\sqrt{3}\right)\right)$$

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.

$$x ^ 2 -6x +6 = 0$$

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

$$r + s = 6 rs = 6$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 6

$$(3 - u) (3 + u) = 6$$

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

$$9 - u^2 = 6$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 6-9 = -3$$

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

$$u^2 = 3 u = \pm\sqrt{3} = \pm \sqrt{3}$$

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

$$r =3 - \sqrt{3} = 1.268 s = 3 + \sqrt{3} = 4.732$$

Evaluate

$$y^{2}-6y+6$$