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r^{2}-\frac{1}{4}r-\frac{1}{4}=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.
r=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\left(-\frac{1}{4}\right)^{2}-4\left(-\frac{1}{4}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{1}{4} for b, and -\frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
r=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{1}{16}-4\left(-\frac{1}{4}\right)}}{2}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
r=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{1}{16}+1}}{2}
Multiply -4 times -\frac{1}{4}.
r=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{17}{16}}}{2}
Add \frac{1}{16} to 1.
r=\frac{-\left(-\frac{1}{4}\right)±\frac{\sqrt{17}}{4}}{2}
Take the square root of \frac{17}{16}.
r=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{2}
The opposite of -\frac{1}{4} is \frac{1}{4}.
r=\frac{\sqrt{17}+1}{2\times 4}
Now solve the equation r=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{2} when ± is plus. Add \frac{1}{4} to \frac{\sqrt{17}}{4}.
r=\frac{\sqrt{17}+1}{8}
Divide \frac{1+\sqrt{17}}{4} by 2.
r=\frac{1-\sqrt{17}}{2\times 4}
Now solve the equation r=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{2} when ± is minus. Subtract \frac{\sqrt{17}}{4} from \frac{1}{4}.
r=\frac{1-\sqrt{17}}{8}
Divide \frac{1-\sqrt{17}}{4} by 2.
r=\frac{\sqrt{17}+1}{8} r=\frac{1-\sqrt{17}}{8}
The equation is now solved.
r^{2}-\frac{1}{4}r-\frac{1}{4}=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.
r^{2}-\frac{1}{4}r-\frac{1}{4}-\left(-\frac{1}{4}\right)=-\left(-\frac{1}{4}\right)
Add \frac{1}{4} to both sides of the equation.
r^{2}-\frac{1}{4}r=-\left(-\frac{1}{4}\right)
Subtracting -\frac{1}{4} from itself leaves 0.
r^{2}-\frac{1}{4}r=\frac{1}{4}
Subtract -\frac{1}{4} from 0.
r^{2}-\frac{1}{4}r+\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.
r^{2}-\frac{1}{4}r+\frac{1}{64}=\frac{1}{4}+\frac{1}{64}
Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.
r^{2}-\frac{1}{4}r+\frac{1}{64}=\frac{17}{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(r-\frac{1}{8}\right)^{2}=\frac{17}{64}
Factor r^{2}-\frac{1}{4}r+\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(r-\frac{1}{8}\right)^{2}}=\sqrt{\frac{17}{64}}
Take the square root of both sides of the equation.
r-\frac{1}{8}=\frac{\sqrt{17}}{8} r-\frac{1}{8}=-\frac{\sqrt{17}}{8}
Simplify.
r=\frac{\sqrt{17}+1}{8} r=\frac{1-\sqrt{17}}{8}
Add \frac{1}{8} to both sides of the equation.