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-64k^{2}+32k+825=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{-32±\sqrt{32^{2}-4\left(-64\right)\times 825}}{2\left(-64\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -64 for a, 32 for b, and 825 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-32±\sqrt{1024-4\left(-64\right)\times 825}}{2\left(-64\right)}
Square 32.
k=\frac{-32±\sqrt{1024+256\times 825}}{2\left(-64\right)}
Multiply -4 times -64.
k=\frac{-32±\sqrt{1024+211200}}{2\left(-64\right)}
Multiply 256 times 825.
k=\frac{-32±\sqrt{212224}}{2\left(-64\right)}
Add 1024 to 211200.
k=\frac{-32±16\sqrt{829}}{2\left(-64\right)}
Take the square root of 212224.
k=\frac{-32±16\sqrt{829}}{-128}
Multiply 2 times -64.
k=\frac{16\sqrt{829}-32}{-128}
Now solve the equation k=\frac{-32±16\sqrt{829}}{-128} when ± is plus. Add -32 to 16\sqrt{829}.
k=-\frac{\sqrt{829}}{8}+\frac{1}{4}
Divide -32+16\sqrt{829} by -128.
k=\frac{-16\sqrt{829}-32}{-128}
Now solve the equation k=\frac{-32±16\sqrt{829}}{-128} when ± is minus. Subtract 16\sqrt{829} from -32.
k=\frac{\sqrt{829}}{8}+\frac{1}{4}
Divide -32-16\sqrt{829} by -128.
k=-\frac{\sqrt{829}}{8}+\frac{1}{4} k=\frac{\sqrt{829}}{8}+\frac{1}{4}
The equation is now solved.
-64k^{2}+32k+825=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.
-64k^{2}+32k+825-825=-825
Subtract 825 from both sides of the equation.
-64k^{2}+32k=-825
Subtracting 825 from itself leaves 0.
\frac{-64k^{2}+32k}{-64}=-\frac{825}{-64}
Divide both sides by -64.
k^{2}+\frac{32}{-64}k=-\frac{825}{-64}
Dividing by -64 undoes the multiplication by -64.
k^{2}-\frac{1}{2}k=-\frac{825}{-64}
Reduce the fraction \frac{32}{-64} to lowest terms by extracting and canceling out 32.
k^{2}-\frac{1}{2}k=\frac{825}{64}
Divide -825 by -64.
k^{2}-\frac{1}{2}k+\left(-\frac{1}{4}\right)^{2}=\frac{825}{64}+\left(-\frac{1}{4}\right)^{2}
Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-\frac{1}{2}k+\frac{1}{16}=\frac{825}{64}+\frac{1}{16}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
k^{2}-\frac{1}{2}k+\frac{1}{16}=\frac{829}{64}
Add \frac{825}{64} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k-\frac{1}{4}\right)^{2}=\frac{829}{64}
Factor k^{2}-\frac{1}{2}k+\frac{1}{16}. 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{1}{4}\right)^{2}}=\sqrt{\frac{829}{64}}
Take the square root of both sides of the equation.
k-\frac{1}{4}=\frac{\sqrt{829}}{8} k-\frac{1}{4}=-\frac{\sqrt{829}}{8}
Simplify.
k=\frac{\sqrt{829}}{8}+\frac{1}{4} k=-\frac{\sqrt{829}}{8}+\frac{1}{4}
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x -\frac{825}{64} = 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.
r + s = \frac{1}{2} rs = -\frac{825}{64}
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}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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}{4} - u) (\frac{1}{4} + u) = -\frac{825}{64}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{825}{64}
\frac{1}{16} - u^2 = -\frac{825}{64}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{825}{64}-\frac{1}{16} = -\frac{829}{64}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = \frac{829}{64} u = \pm\sqrt{\frac{829}{64}} = \pm \frac{\sqrt{829}}{8}
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}{4} - \frac{\sqrt{829}}{8} = -3.349 s = \frac{1}{4} + \frac{\sqrt{829}}{8} = 3.849
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.