Question

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

• A. $x=-1$ and $x=\frac{2}{3}$
• B. $x=1$ and $x=\frac{1}{3}$
• C. $x=1$ and $x=\frac{2}{3}$
• D. None of these

Answer 1

Answer: (C)

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

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

$\Rightarrow (3x{)}^2-(3x\left)\right(5)+6=0$

$\Rightarrow (3x{)}^2-(3x)2\left(\frac{5}{2}\right)+6=0$

$\Rightarrow \left[\right(3x{)}^2-\left(3x\right)\times \left(2\right)\times \left(\frac{5}{2}\right)+{\left(\frac{5}{2}\right)}^2]-{\left(\frac{5}{2}\right)}^2+6=0$

$\Rightarrow {(3x-\frac{5}{2})}^2-\frac{25}{4}+6=0$

$\Rightarrow {(3x-\frac{5}{2})}^2=\frac{25}{4}-6$

$\Rightarrow {(3x-\frac{5}{2})}^2=\frac{1}{4}$

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

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

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

$\Rightarrow x=\frac{2}{3}$ or $x=1$

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