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64x^{2}-16x+16=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(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 64\times 16}}{2\times 64}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, -16 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-16\right)±\sqrt{256-4\times 64\times 16}}{2\times 64}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256-256\times 16}}{2\times 64}
Multiply -4 times 64.
x=\frac{-\left(-16\right)±\sqrt{256-4096}}{2\times 64}
Multiply -256 times 16.
x=\frac{-\left(-16\right)±\sqrt{-3840}}{2\times 64}
Add 256 to -4096.
x=\frac{-\left(-16\right)±16\sqrt{15}i}{2\times 64}
Take the square root of -3840.
x=\frac{16±16\sqrt{15}i}{2\times 64}
The opposite of -16 is 16.
x=\frac{16±16\sqrt{15}i}{128}
Multiply 2 times 64.
x=\frac{16+16\sqrt{15}i}{128}
Now solve the equation x=\frac{16±16\sqrt{15}i}{128} when ± is plus. Add 16 to 16i\sqrt{15}.
x=\frac{1+\sqrt{15}i}{8}
Divide 16+16i\sqrt{15} by 128.
x=\frac{-16\sqrt{15}i+16}{128}
Now solve the equation x=\frac{16±16\sqrt{15}i}{128} when ± is minus. Subtract 16i\sqrt{15} from 16.
x=\frac{-\sqrt{15}i+1}{8}
Divide 16-16i\sqrt{15} by 128.
x=\frac{1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i+1}{8}
The equation is now solved.
64x^{2}-16x+16=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.
64x^{2}-16x+16-16=-16
Subtract 16 from both sides of the equation.
64x^{2}-16x=-16
Subtracting 16 from itself leaves 0.
\frac{64x^{2}-16x}{64}=-\frac{16}{64}
Divide both sides by 64.
x^{2}+\left(-\frac{16}{64}\right)x=-\frac{16}{64}
Dividing by 64 undoes the multiplication by 64.
x^{2}-\frac{1}{4}x=-\frac{16}{64}
Reduce the fraction \frac{-16}{64} to lowest terms by extracting and canceling out 16.
x^{2}-\frac{1}{4}x=-\frac{1}{4}
Reduce the fraction \frac{-16}{64} to lowest terms by extracting and canceling out 16.
x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=-\frac{1}{4}+\left(-\frac{1}{8}\right)^{2}
Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{1}{4}+\frac{1}{64}
Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{15}{64}
Add -\frac{1}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{8}\right)^{2}=-\frac{15}{64}
Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. 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}{8}\right)^{2}}=\sqrt{-\frac{15}{64}}
Take the square root of both sides of the equation.
x-\frac{1}{8}=\frac{\sqrt{15}i}{8} x-\frac{1}{8}=-\frac{\sqrt{15}i}{8}
Simplify.
x=\frac{1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i+1}{8}
Add \frac{1}{8} to both sides of the equation.
x ^ 2 -\frac{1}{4}x +\frac{1}{4} = 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 64
r + s = \frac{1}{4} rs = \frac{1}{4}
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}{8} - u s = \frac{1}{8} + u
Two numbers r and s sum up to \frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{4} = \frac{1}{8}. 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}{8} - u) (\frac{1}{8} + u) = \frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}
\frac{1}{64} - u^2 = \frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{4}-\frac{1}{64} = \frac{15}{64}
Simplify the expression by subtracting \frac{1}{64} on both sides
u^2 = -\frac{15}{64} u = \pm\sqrt{-\frac{15}{64}} = \pm \frac{\sqrt{15}}{8}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{1}{8} - \frac{\sqrt{15}}{8}i = 0.125 - 0.484i s = \frac{1}{8} + \frac{\sqrt{15}}{8}i = 0.125 + 0.484i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.