Solve for x
x=\frac{\sqrt{3}+1}{20}\approx 0.13660254
x=\frac{1-\sqrt{3}}{20}\approx -0.03660254
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200x^{2}-20x-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(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 200\left(-1\right)}}{2\times 200}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 200 for a, -20 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-20\right)±\sqrt{400-4\times 200\left(-1\right)}}{2\times 200}
Square -20.
x=\frac{-\left(-20\right)±\sqrt{400-800\left(-1\right)}}{2\times 200}
Multiply -4 times 200.
x=\frac{-\left(-20\right)±\sqrt{400+800}}{2\times 200}
Multiply -800 times -1.
x=\frac{-\left(-20\right)±\sqrt{1200}}{2\times 200}
Add 400 to 800.
x=\frac{-\left(-20\right)±20\sqrt{3}}{2\times 200}
Take the square root of 1200.
x=\frac{20±20\sqrt{3}}{2\times 200}
The opposite of -20 is 20.
x=\frac{20±20\sqrt{3}}{400}
Multiply 2 times 200.
x=\frac{20\sqrt{3}+20}{400}
Now solve the equation x=\frac{20±20\sqrt{3}}{400} when ± is plus. Add 20 to 20\sqrt{3}.
x=\frac{\sqrt{3}+1}{20}
Divide 20+20\sqrt{3} by 400.
x=\frac{20-20\sqrt{3}}{400}
Now solve the equation x=\frac{20±20\sqrt{3}}{400} when ± is minus. Subtract 20\sqrt{3} from 20.
x=\frac{1-\sqrt{3}}{20}
Divide 20-20\sqrt{3} by 400.
x=\frac{\sqrt{3}+1}{20} x=\frac{1-\sqrt{3}}{20}
The equation is now solved.
200x^{2}-20x-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.
200x^{2}-20x-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
200x^{2}-20x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
200x^{2}-20x=1
Subtract -1 from 0.
\frac{200x^{2}-20x}{200}=\frac{1}{200}
Divide both sides by 200.
x^{2}+\left(-\frac{20}{200}\right)x=\frac{1}{200}
Dividing by 200 undoes the multiplication by 200.
x^{2}-\frac{1}{10}x=\frac{1}{200}
Reduce the fraction \frac{-20}{200} to lowest terms by extracting and canceling out 20.
x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\frac{1}{200}+\left(-\frac{1}{20}\right)^{2}
Divide -\frac{1}{10}, the coefficient of the x term, by 2 to get -\frac{1}{20}. Then add the square of -\frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{1}{200}+\frac{1}{400}
Square -\frac{1}{20} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{3}{400}
Add \frac{1}{200} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{20}\right)^{2}=\frac{3}{400}
Factor x^{2}-\frac{1}{10}x+\frac{1}{400}. 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}{20}\right)^{2}}=\sqrt{\frac{3}{400}}
Take the square root of both sides of the equation.
x-\frac{1}{20}=\frac{\sqrt{3}}{20} x-\frac{1}{20}=-\frac{\sqrt{3}}{20}
Simplify.
x=\frac{\sqrt{3}+1}{20} x=\frac{1-\sqrt{3}}{20}
Add \frac{1}{20} to both sides of the equation.
x ^ 2 -\frac{1}{10}x -\frac{1}{200} = 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 200
r + s = \frac{1}{10} rs = -\frac{1}{200}
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}{20} - u s = \frac{1}{20} + u
Two numbers r and s sum up to \frac{1}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{10} = \frac{1}{20}. 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}{20} - u) (\frac{1}{20} + u) = -\frac{1}{200}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{200}
\frac{1}{400} - u^2 = -\frac{1}{200}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{200}-\frac{1}{400} = -\frac{3}{400}
Simplify the expression by subtracting \frac{1}{400} on both sides
u^2 = \frac{3}{400} u = \pm\sqrt{\frac{3}{400}} = \pm \frac{\sqrt{3}}{20}
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}{20} - \frac{\sqrt{3}}{20} = -0.037 s = \frac{1}{20} + \frac{\sqrt{3}}{20} = 0.137
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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