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8x^{2}-5x+500=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 8\times 500}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -5 for b, and 500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 8\times 500}}{2\times 8}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-32\times 500}}{2\times 8}
Multiply -4 times 8.
x=\frac{-\left(-5\right)±\sqrt{25-16000}}{2\times 8}
Multiply -32 times 500.
x=\frac{-\left(-5\right)±\sqrt{-15975}}{2\times 8}
Add 25 to -16000.
x=\frac{-\left(-5\right)±15\sqrt{71}i}{2\times 8}
Take the square root of -15975.
x=\frac{5±15\sqrt{71}i}{2\times 8}
The opposite of -5 is 5.
x=\frac{5±15\sqrt{71}i}{16}
Multiply 2 times 8.
x=\frac{5+15\sqrt{71}i}{16}
Now solve the equation x=\frac{5±15\sqrt{71}i}{16} when ± is plus. Add 5 to 15i\sqrt{71}.
x=\frac{-15\sqrt{71}i+5}{16}
Now solve the equation x=\frac{5±15\sqrt{71}i}{16} when ± is minus. Subtract 15i\sqrt{71} from 5.
x=\frac{5+15\sqrt{71}i}{16} x=\frac{-15\sqrt{71}i+5}{16}
The equation is now solved.
8x^{2}-5x+500=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.
8x^{2}-5x+500-500=-500
Subtract 500 from both sides of the equation.
8x^{2}-5x=-500
Subtracting 500 from itself leaves 0.
\frac{8x^{2}-5x}{8}=-\frac{500}{8}
Divide both sides by 8.
x^{2}-\frac{5}{8}x=-\frac{500}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}-\frac{5}{8}x=-\frac{125}{2}
Reduce the fraction \frac{-500}{8} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{125}{2}+\left(-\frac{5}{16}\right)^{2}
Divide -\frac{5}{8}, the coefficient of the x term, by 2 to get -\frac{5}{16}. Then add the square of -\frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{125}{2}+\frac{25}{256}
Square -\frac{5}{16} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{15975}{256}
Add -\frac{125}{2} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{16}\right)^{2}=-\frac{15975}{256}
Factor x^{2}-\frac{5}{8}x+\frac{25}{256}. 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{5}{16}\right)^{2}}=\sqrt{-\frac{15975}{256}}
Take the square root of both sides of the equation.
x-\frac{5}{16}=\frac{15\sqrt{71}i}{16} x-\frac{5}{16}=-\frac{15\sqrt{71}i}{16}
Simplify.
x=\frac{5+15\sqrt{71}i}{16} x=\frac{-15\sqrt{71}i+5}{16}
Add \frac{5}{16} to both sides of the equation.
x ^ 2 -\frac{5}{8}x +\frac{125}{2} = 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 8
r + s = \frac{5}{8} rs = \frac{125}{2}
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{5}{16} - u s = \frac{5}{16} + u
Two numbers r and s sum up to \frac{5}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{8} = \frac{5}{16}. 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{5}{16} - u) (\frac{5}{16} + u) = \frac{125}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{125}{2}
\frac{25}{256} - u^2 = \frac{125}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{125}{2}-\frac{25}{256} = \frac{15975}{256}
Simplify the expression by subtracting \frac{25}{256} on both sides
u^2 = -\frac{15975}{256} u = \pm\sqrt{-\frac{15975}{256}} = \pm \frac{\sqrt{15975}}{16}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{5}{16} - \frac{\sqrt{15975}}{16}i = 0.313 - 7.900i s = \frac{5}{16} + \frac{\sqrt{15975}}{16}i = 0.313 + 7.900i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.