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5k^{2}-4k+8=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.
k=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\times 8}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-4\right)±\sqrt{16-4\times 5\times 8}}{2\times 5}
Square -4.
k=\frac{-\left(-4\right)±\sqrt{16-20\times 8}}{2\times 5}
Multiply -4 times 5.
k=\frac{-\left(-4\right)±\sqrt{16-160}}{2\times 5}
Multiply -20 times 8.
k=\frac{-\left(-4\right)±\sqrt{-144}}{2\times 5}
Add 16 to -160.
k=\frac{-\left(-4\right)±12i}{2\times 5}
Take the square root of -144.
k=\frac{4±12i}{2\times 5}
The opposite of -4 is 4.
k=\frac{4±12i}{10}
Multiply 2 times 5.
k=\frac{4+12i}{10}
Now solve the equation k=\frac{4±12i}{10} when ± is plus. Add 4 to 12i.
k=\frac{2}{5}+\frac{6}{5}i
Divide 4+12i by 10.
k=\frac{4-12i}{10}
Now solve the equation k=\frac{4±12i}{10} when ± is minus. Subtract 12i from 4.
k=\frac{2}{5}-\frac{6}{5}i
Divide 4-12i by 10.
k=\frac{2}{5}+\frac{6}{5}i k=\frac{2}{5}-\frac{6}{5}i
The equation is now solved.
5k^{2}-4k+8=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.
5k^{2}-4k+8-8=-8
Subtract 8 from both sides of the equation.
5k^{2}-4k=-8
Subtracting 8 from itself leaves 0.
\frac{5k^{2}-4k}{5}=-\frac{8}{5}
Divide both sides by 5.
k^{2}-\frac{4}{5}k=-\frac{8}{5}
Dividing by 5 undoes the multiplication by 5.
k^{2}-\frac{4}{5}k+\left(-\frac{2}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{2}{5}\right)^{2}
Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-\frac{4}{5}k+\frac{4}{25}=-\frac{8}{5}+\frac{4}{25}
Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.
k^{2}-\frac{4}{5}k+\frac{4}{25}=-\frac{36}{25}
Add -\frac{8}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k-\frac{2}{5}\right)^{2}=-\frac{36}{25}
Factor k^{2}-\frac{4}{5}k+\frac{4}{25}. 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(k-\frac{2}{5}\right)^{2}}=\sqrt{-\frac{36}{25}}
Take the square root of both sides of the equation.
k-\frac{2}{5}=\frac{6}{5}i k-\frac{2}{5}=-\frac{6}{5}i
Simplify.
k=\frac{2}{5}+\frac{6}{5}i k=\frac{2}{5}-\frac{6}{5}i
Add \frac{2}{5} to both sides of the equation.
x ^ 2 -\frac{4}{5}x +\frac{8}{5} = 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 5
r + s = \frac{4}{5} rs = \frac{8}{5}
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{2}{5} - u s = \frac{2}{5} + u
Two numbers r and s sum up to \frac{4}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{5} = \frac{2}{5}. 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{2}{5} - u) (\frac{2}{5} + u) = \frac{8}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{5}
\frac{4}{25} - u^2 = \frac{8}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{8}{5}-\frac{4}{25} = \frac{36}{25}
Simplify the expression by subtracting \frac{4}{25} on both sides
u^2 = -\frac{36}{25} u = \pm\sqrt{-\frac{36}{25}} = \pm \frac{6}{5}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{2}{5} - \frac{6}{5}i = 0.400 - 1.200i s = \frac{2}{5} + \frac{6}{5}i = 0.400 + 1.200i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.