Question

Solve the following by using the method of completing square.
$6x^2-11x+3=0$

Answer 1

We have,

$\begin{array}{c}6x^2-11x+3=0\\ \Rightarrow 6\left[x^2-\frac{11}{6}x+{\left(\frac{11}{12}\right)}^2-{\left(\frac{11}{12}\right)}^2\right]+3=0\\ \Rightarrow 6{\left(x-\frac{11}{12}\right)}^2-6\times \frac{121}{144}+3=0\\ \Rightarrow 6{\left(x-\frac{11}{12}\right)}^2=\frac{121}{24}-3\\ \Rightarrow 6{\left(x-\frac{11}{12}\right)}^2=\frac{49}{24}\\ \Rightarrow {\left(x-\frac{11}{12}\right)}^2=\frac{49}{144}\\ \Rightarrow x-\frac{11}{12}=\pm \frac{7}{12}\\ \therefore x=\frac{3}{2}orx=\frac{1}{3}\end{array}$