Solve 5x^2-5x+1

$$5 x ^ { 2 } - 5 x + 1$$

Factor

$$5\left(x-\left(-\frac{\sqrt{5}}{10}+\frac{1}{2}\right)\right)\left(x-\left(\frac{\sqrt{5}}{10}+\frac{1}{2}\right)\right)$$

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.

$$5x^{2}-5x+1=0$$

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.

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

Square -5.

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

Multiply -4 times 5.

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

Add 25 to -20.

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

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{5}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{5±\sqrt{5}}{10}$$

Now solve the equation x=\frac{5±\sqrt{5}}{10} when ± is plus. Add 5 to \sqrt{5}.

$$x=\frac{\sqrt{5}+5}{10}$$

Divide 5+\sqrt{5} by 10.

$$x=\frac{\sqrt{5}}{10}+\frac{1}{2}$$

Now solve the equation x=\frac{5±\sqrt{5}}{10} when ± is minus. Subtract \sqrt{5} from 5.

$$x=\frac{5-\sqrt{5}}{10}$$

Divide 5-\sqrt{5} by 10.

$$x=-\frac{\sqrt{5}}{10}+\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2}+\frac{\sqrt{5}}{10} for x_{1} and \frac{1}{2}-\frac{\sqrt{5}}{10} for x_{2}.

$$5x^{2}-5x+1=5\left(x-\left(\frac{\sqrt{5}}{10}+\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{5}}{10}+\frac{1}{2}\right)\right)$$

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.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 -1x +\frac{1}{5} = 0$$

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

$$r + s = 1 rs = \frac{1}{5}$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = \frac{1}{5}$$

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

$$\frac{1}{4} - u^2 = \frac{1}{5}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = \frac{1}{5}-\frac{1}{4} = -\frac{1}{20}$$

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

$$u^2 = \frac{1}{20} u = \pm\sqrt{\frac{1}{20}} = \pm \frac{1}{\sqrt{20}}$$

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

$$r =\frac{1}{2} - \frac{1}{\sqrt{20}} = 0.276 s = \frac{1}{2} + \frac{1}{\sqrt{20}} = 0.724$$

Evaluate

$$5x^{2}-5x+1$$