Solve f(x)=x^2-5x+5

$$f ( x ) = x ^ { 2 } - 5 x + 5$$

Factor

$$\left(x-\frac{5-\sqrt{5}}{2}\right)\left(x-\frac{\sqrt{5}+5}{2}\right)$$

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.

$$x^{2}-5x+5=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-20}}{2}$$

Add 25 to -20.

$$x=\frac{-\left(-5\right)±\sqrt{5}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}+5}{2}$$

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

$$x=\frac{5-\sqrt{5}}{2}$$

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

$$x^{2}-5x+5=\left(x-\frac{\sqrt{5}+5}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)$$

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.

$$x ^ 2 -5x +5 = 0$$

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

$$r + s = 5 rs = 5$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = 5$$

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

$$\frac{25}{4} - u^2 = 5$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = 5-\frac{25}{4} = -\frac{5}{4}$$

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

$$u^2 = \frac{5}{4} u = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$$

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

$$r =\frac{5}{2} - \frac{\sqrt{5}}{2} = 1.382 s = \frac{5}{2} + \frac{\sqrt{5}}{2} = 3.618$$

Evaluate

$$x^{2}-5x+5$$