Solve 5x^2+2x-1

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

Factor

$$5\left(x-\frac{-\sqrt{6}-1}{5}\right)\left(x-\frac{\sqrt{6}-1}{5}\right)$$

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.

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

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-2±\sqrt{4+20}}{2\times 5}$$

Add 4 to 20.

$$x=\frac{-2±\sqrt{24}}{2\times 5}$$

Take the square root of 24.

$$x=\frac{-2±2\sqrt{6}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±2\sqrt{6}}{10}$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{10} when ± is plus. Add -2 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-2}{10}$$

Divide -2+2\sqrt{6} by 10.

$$x=\frac{\sqrt{6}-1}{5}$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{10} when ± is minus. Subtract 2\sqrt{6} from -2.

$$x=\frac{-2\sqrt{6}-2}{10}$$

Divide -2-2\sqrt{6} by 10.

$$x=\frac{-\sqrt{6}-1}{5}$$

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

$$5x^{2}+2x-1=5\left(x-\frac{\sqrt{6}-1}{5}\right)\left(x-\frac{-\sqrt{6}-1}{5}\right)$$

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

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

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

$$r + s = -\frac{2}{5} rs = -\frac{1}{5}$$

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

$$r = -\frac{1}{5} - u s = -\frac{1}{5} + u$$

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

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

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

$$\frac{1}{25} - u^2 = -\frac{1}{5}$$

Simplify the expression by subtracting \frac{1}{25} on both sides

$$-u^2 = -\frac{1}{5}-\frac{1}{25} = -\frac{6}{25}$$

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

$$u^2 = \frac{6}{25} u = \pm\sqrt{\frac{6}{25}} = \pm \frac{\sqrt{6}}{5}$$

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

$$r =-\frac{1}{5} - \frac{\sqrt{6}}{5} = -0.690 s = -\frac{1}{5} + \frac{\sqrt{6}}{5} = 0.290$$

Evaluate

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