Question

Solve the quadratic equation $9x^2-15x+6=0$ by the method of completing the square.

Answer 1

We have
$9x^2-15x+6=0$

$\Rightarrow x^2-\frac{15}{9}x+\frac{6}{9}=0$ (Dividing throughout by $9$)

$\Rightarrow x^2-\frac{5}{3}x+\frac{2}{3}=0\Rightarrow x^2-\frac{5}{3}x=-\frac{2}{3}$ (Shifting the constant term on RHS)

$\Rightarrow x^2-2\left(\frac{5}{6}\right)x+{\left(\frac{5}{6}\right)}^2={\left(\frac{5}{6}\right)}^2-\frac{2}{3}$ (Adding square of half of coefficient of $x$ on both sides)

$\Rightarrow {(x-\frac{5}{6})}^2=\frac{25}{36}-\frac{2}{3}\Rightarrow {(x-\frac{5}{6})}^2=\frac{1}{36}\Rightarrow x-\frac{5}{6}=\pm \frac{1}{6}$ (Taking square root on both sides)

$\Rightarrow x=\frac{5}{6}+\frac{1}{6}=1orx=\frac{5}{6}-\frac{1}{6}=\frac{2}{3}\Rightarrow x=1orx=\frac{2}{3}$

Hence the roots of the equation are $1$ and $\frac{2}{3}$