Solve for x (complex solution)
x=\frac{3+\sqrt{115}i}{62}\approx 0.048387097+0.172964602i
x=\frac{-\sqrt{115}i+3}{62}\approx 0.048387097-0.172964602i
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31x^{2}-3x+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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 31}}{2\times 31}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 31 for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 31}}{2\times 31}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-124}}{2\times 31}
Multiply -4 times 31.
x=\frac{-\left(-3\right)±\sqrt{-115}}{2\times 31}
Add 9 to -124.
x=\frac{-\left(-3\right)±\sqrt{115}i}{2\times 31}
Take the square root of -115.
x=\frac{3±\sqrt{115}i}{2\times 31}
The opposite of -3 is 3.
x=\frac{3±\sqrt{115}i}{62}
Multiply 2 times 31.
x=\frac{3+\sqrt{115}i}{62}
Now solve the equation x=\frac{3±\sqrt{115}i}{62} when ± is plus. Add 3 to i\sqrt{115}.
x=\frac{-\sqrt{115}i+3}{62}
Now solve the equation x=\frac{3±\sqrt{115}i}{62} when ± is minus. Subtract i\sqrt{115} from 3.
x=\frac{3+\sqrt{115}i}{62} x=\frac{-\sqrt{115}i+3}{62}
The equation is now solved.
31x^{2}-3x+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.
31x^{2}-3x+1-1=-1
Subtract 1 from both sides of the equation.
31x^{2}-3x=-1
Subtracting 1 from itself leaves 0.
\frac{31x^{2}-3x}{31}=-\frac{1}{31}
Divide both sides by 31.
x^{2}-\frac{3}{31}x=-\frac{1}{31}
Dividing by 31 undoes the multiplication by 31.
x^{2}-\frac{3}{31}x+\left(-\frac{3}{62}\right)^{2}=-\frac{1}{31}+\left(-\frac{3}{62}\right)^{2}
Divide -\frac{3}{31}, the coefficient of the x term, by 2 to get -\frac{3}{62}. Then add the square of -\frac{3}{62} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{31}x+\frac{9}{3844}=-\frac{1}{31}+\frac{9}{3844}
Square -\frac{3}{62} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{31}x+\frac{9}{3844}=-\frac{115}{3844}
Add -\frac{1}{31} to \frac{9}{3844} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{62}\right)^{2}=-\frac{115}{3844}
Factor x^{2}-\frac{3}{31}x+\frac{9}{3844}. 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{3}{62}\right)^{2}}=\sqrt{-\frac{115}{3844}}
Take the square root of both sides of the equation.
x-\frac{3}{62}=\frac{\sqrt{115}i}{62} x-\frac{3}{62}=-\frac{\sqrt{115}i}{62}
Simplify.
x=\frac{3+\sqrt{115}i}{62} x=\frac{-\sqrt{115}i+3}{62}
Add \frac{3}{62} to both sides of the equation.
x ^ 2 -\frac{3}{31}x +\frac{1}{31} = 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 31
r + s = \frac{3}{31} rs = \frac{1}{31}
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{3}{62} - u s = \frac{3}{62} + u
Two numbers r and s sum up to \frac{3}{31} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{31} = \frac{3}{62}. 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{3}{62} - u) (\frac{3}{62} + u) = \frac{1}{31}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{31}
\frac{9}{3844} - u^2 = \frac{1}{31}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{31}-\frac{9}{3844} = \frac{115}{3844}
Simplify the expression by subtracting \frac{9}{3844} on both sides
u^2 = -\frac{115}{3844} u = \pm\sqrt{-\frac{115}{3844}} = \pm \frac{\sqrt{115}}{62}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{62} - \frac{\sqrt{115}}{62}i = 0.048 - 0.173i s = \frac{3}{62} + \frac{\sqrt{115}}{62}i = 0.048 + 0.173i
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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