Question

\(2x^2-6x+1=0\)

Answer 1

\(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{7}{2}=0\)

\(\Leftrightarrow2\left(x-\dfrac{3}{2}\right)^2=\dfrac{7}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\sqrt{\dfrac{7}{4}}\\x-\dfrac{3}{2}=-\sqrt{\dfrac{7}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{7}{4}}+\dfrac{3}{2}\\x=-\sqrt{\dfrac{7}{4}}+\dfrac{3}{2}\end{matrix}\right.\)