Solve 27x^2+18x+1=

$$27 x ^ { 2 } + 18 x + 1 =$$

Factor

$$27\left(x-\left(-\frac{\sqrt{6}}{9}-\frac{1}{3}\right)\right)\left(x-\left(\frac{\sqrt{6}}{9}-\frac{1}{3}\right)\right)$$

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.

$$27x^{2}+18x+1=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 27}}{2\times 27}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 27}}{2\times 27}$$

Multiply -4 times 27.

$$x=\frac{-18±\sqrt{324-108}}{2\times 27}$$

Add 324 to -108.

$$x=\frac{-18±\sqrt{216}}{2\times 27}$$

Take the square root of 216.

$$x=\frac{-18±6\sqrt{6}}{2\times 27}$$

Multiply 2 times 27.

$$x=\frac{-18±6\sqrt{6}}{54}$$

Now solve the equation x=\frac{-18±6\sqrt{6}}{54} when ± is plus. Add -18 to 6\sqrt{6}.

$$x=\frac{6\sqrt{6}-18}{54}$$

Divide -18+6\sqrt{6} by 54.

$$x=\frac{\sqrt{6}}{9}-\frac{1}{3}$$

Now solve the equation x=\frac{-18±6\sqrt{6}}{54} when ± is minus. Subtract 6\sqrt{6} from -18.

$$x=\frac{-6\sqrt{6}-18}{54}$$

Divide -18-6\sqrt{6} by 54.

$$x=-\frac{\sqrt{6}}{9}-\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3}+\frac{\sqrt{6}}{9} for x_{1} and -\frac{1}{3}-\frac{\sqrt{6}}{9} for x_{2}.

$$27x^{2}+18x+1=27\left(x-\left(\frac{\sqrt{6}}{9}-\frac{1}{3}\right)\right)\left(x-\left(-\frac{\sqrt{6}}{9}-\frac{1}{3}\right)\right)$$

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.This is achieved by dividing both sides of the equation by 27

$$x ^ 2 +\frac{2}{3}x +\frac{1}{27} = 0$$

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

$$r + s = -\frac{2}{3} rs = \frac{1}{27}$$

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

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{27}

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = \frac{1}{27}$$

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

$$\frac{1}{9} - u^2 = \frac{1}{27}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = \frac{1}{27}-\frac{1}{9} = -\frac{2}{27}$$

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

$$u^2 = \frac{2}{27} u = \pm\sqrt{\frac{2}{27}} = \pm \frac{\sqrt{2}}{\sqrt{27}}$$

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

$$r =-\frac{1}{3} - \frac{\sqrt{2}}{\sqrt{27}} = -0.605 s = -\frac{1}{3} + \frac{\sqrt{2}}{\sqrt{27}} = -0.061$$

Evaluate

$$27x^{2}+18x+1$$