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k\left(k-1\right)=0
Factor out k.
k=0 k=1
To find equation solutions, solve k=0 and k-1=0.
k^{2}-k=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(-1\right)±\sqrt{1}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-1\right)±1}{2}
Take the square root of 1.
k=\frac{1±1}{2}
The opposite of -1 is 1.
k=\frac{2}{2}
Now solve the equation k=\frac{1±1}{2} when ± is plus. Add 1 to 1.
k=1
Divide 2 by 2.
k=\frac{0}{2}
Now solve the equation k=\frac{1±1}{2} when ± is minus. Subtract 1 from 1.
k=0
Divide 0 by 2.
k=1 k=0
The equation is now solved.
k^{2}-k=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.
k^{2}-k+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-k+\frac{1}{4}=\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
\left(k-\frac{1}{2}\right)^{2}=\frac{1}{4}
Factor k^{2}-k+\frac{1}{4}. 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}{2}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
k-\frac{1}{2}=\frac{1}{2} k-\frac{1}{2}=-\frac{1}{2}
Simplify.
k=1 k=0
Add \frac{1}{2} to both sides of the equation.