Question

Solve $3x^2-5x+2=0$ by completing the square method.

Answer 1

Given $3x^2-5x+2=0$

Here, the coefficient of $x^2$ is 3 and it is not a perfect square.

$\therefore$ Multiply the equation throughout by 3.

$\Rightarrow (3x^2-5x+2=0)\times 3\therefore 9x^2-15x+6=0$

Now, half of the coefficient of x is $\frac{5}{2}.\therefore b=\frac{5}{2}$ and $b^2={\left(\frac{5}{2}\right)}^2$

So, $9x^2-15x+{\left(\frac{5}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2+6=0,(3x{)}^2-15x+{\left(\frac{5}{2}\right)}^2={\left(\frac{5}{2}\right)}^2-6$

${(3x-\frac{5}{2})}^2=\frac{25}{4}-6=\frac{1}{4}$
(Taking square root on both the sides)

$\therefore 3x-\frac{5}{2}=\pm \frac{1}{2}$

$\therefore 3x-\frac{5}{2}=\pm \frac{1}{2}$ or $3x-\frac{5}{2}=-\frac{1}{2}$

$3x=\frac{1}{2}+\frac{5}{2}$ $3x=\frac{5}{2}-\frac{1}{2}$

$3x=\frac{6}{2}=3$ $3x=\frac{4}{2}=2$

$\therefore x=1$ $x=\frac{2}{3}$