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Question
solve the quadratic equation by completing the square. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.) 9x^2 - 18x = - 6
Step1: Make coefficient of $x^{2}$ equal 1
Divide the entire equation $9x^{2}-18x = - 6$ by 9. We get $x^{2}-2x=-\frac{6}{9}=-\frac{2}{3}$.
Step2: Complete the square
Take half of the coefficient of $x$ (which is - 2), square it. Half of -2 is -1, and $(-1)^{2}=1$. Add 1 to both sides of the equation: $x^{2}-2x + 1=-\frac{2}{3}+1$. The left - hand side is a perfect square $(x - 1)^{2}$, and the right - hand side is $-\frac{2}{3}+1=\frac{-2 + 3}{3}=\frac{1}{3}$. So, $(x - 1)^{2}=\frac{1}{3}$.
Step3: Solve for $x$
Take the square root of both sides: $x-1=\pm\sqrt{\frac{1}{3}}=\pm\frac{\sqrt{3}}{3}$. Then $x = 1\pm\frac{\sqrt{3}}{3}$.
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$1+\frac{\sqrt{3}}{3},1-\frac{\sqrt{3}}{3}$