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6x^{2}-6x+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(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 6}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\times 6}}{2\times 6}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36-24}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-6\right)±\sqrt{12}}{2\times 6}
Add 36 to -24.
x=\frac{-\left(-6\right)±2\sqrt{3}}{2\times 6}
Take the square root of 12.
x=\frac{6±2\sqrt{3}}{2\times 6}
The opposite of -6 is 6.
x=\frac{6±2\sqrt{3}}{12}
Multiply 2 times 6.
x=\frac{2\sqrt{3}+6}{12}
Now solve the equation x=\frac{6±2\sqrt{3}}{12} when ± is plus. Add 6 to 2\sqrt{3}.
x=\frac{\sqrt{3}}{6}+\frac{1}{2}
Divide 6+2\sqrt{3} by 12.
x=\frac{6-2\sqrt{3}}{12}
Now solve the equation x=\frac{6±2\sqrt{3}}{12} when ± is minus. Subtract 2\sqrt{3} from 6.
x=-\frac{\sqrt{3}}{6}+\frac{1}{2}
Divide 6-2\sqrt{3} by 12.
x=\frac{\sqrt{3}}{6}+\frac{1}{2} x=-\frac{\sqrt{3}}{6}+\frac{1}{2}
The equation is now solved.
6x^{2}-6x+1=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
6x^{2}-6x+1-1=-1
Subtract 1 from both sides of the equation.
6x^{2}-6x=-1
Subtracting 1 from itself leaves 0.
\frac{6x^{2}-6x}{6}=-\frac{1}{6}
Divide both sides by 6.
x^{2}+\left(-\frac{6}{6}\right)x=-\frac{1}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-x=-\frac{1}{6}
Divide -6 by 6.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{6}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=-\frac{1}{6}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{1}{12}
Add -\frac{1}{6} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=\frac{1}{12}
Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{12}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{3}}{6} x-\frac{1}{2}=-\frac{\sqrt{3}}{6}
Simplify.
x=\frac{\sqrt{3}}{6}+\frac{1}{2} x=-\frac{\sqrt{3}}{6}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x +\frac{1}{6} = 0
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 6
r + s = 1 rs = \frac{1}{6}
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 = \frac{1}{2} - u s = \frac{1}{2} + u
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>
(\frac{1}{2} - u) (\frac{1}{2} + u) = \frac{1}{6}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{6}
\frac{1}{4} - u^2 = \frac{1}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{6}-\frac{1}{4} = -\frac{1}{12}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{1}{12} u = \pm\sqrt{\frac{1}{12}} = \pm \frac{1}{\sqrt{12}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - \frac{1}{\sqrt{12}} = 0.211 s = \frac{1}{2} + \frac{1}{\sqrt{12}} = 0.789
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.