Question

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

Answer 1

Hint: This method is always used for solving quadratic equations. In this method We first make the coefficient of $ {x^2} $ as 1 by dividing the complete equation by coefficient of $ {x^2} $ then add the square of half of the coefficient of ‘x’ on both the side of the equation. After that we shift the constant term on the right side of the equation then we simplify the left side of the equation by using some algebraic identity and solve the right side of the equation by taking LCM then by taking the square root on both sides we get the required roots.

Complete step-by-step answer:
Given quadratic equation $ 3{x^2} - 5x + 2 = 0 $
For making the coefficient of $ {x^2} $ as 1 we divide the whole equation by 3
 $ $ \[{x^2} - \dfrac{5}{3}x + \dfrac{2}{3} = 0\]
Then we shift the constant term $ \dfrac{2}{3} $ on the right side of equation
\[{x^2} - \dfrac{5}{3}x = - \dfrac{2}{3}\] ------- $ \left( 1 \right) $
Now the coefficient of ‘x’ is $ - \dfrac{5}{3} $ .for getting the half of $ - \dfrac{5}{3} $ we multiply it by $ \dfrac{1}{2} $ then we square it $ $
\[{\left( { - \dfrac{5}{3} \times \dfrac{1}{2}} \right)^2} = {\left( { - \dfrac{5}{6}} \right)^2} = \dfrac{{25}}{{36}}\]
Now we need to add $ \dfrac{{25}}{{36}} $ on both side of equation (1)
\[{x^2} - \dfrac{5}{3}x + \dfrac{{25}}{{36}} = - \dfrac{2}{3} + \dfrac{{25}}{{36}}\]
By comparing the left side of the equation with algebraic identity $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $
We get $ a = x $ and $ b = \dfrac{5}{6} $ also solve the right side of the equation by taking 36 as LCM of 3 and 36.
 $ {\left( {x - \dfrac{5}{6}} \right)^2} = \dfrac{{ - 24 + 25}}{{36}} $
 $ \Rightarrow {\left( {x - \dfrac{5}{6}} \right)^2} = \dfrac{1}{{36}} $
Now by taking square root on both on both side of the equation
 $ \Rightarrow x - \dfrac{5}{6} = \pm \sqrt {\dfrac{1}{{36}}} $
Since the square root of $ \dfrac{1}{{36}} $ is $ \dfrac{1}{6} $ then above equation can written as
 $ \Rightarrow x - \dfrac{5}{6} = \pm \dfrac{1}{6} $ --------- $ \left( 2 \right) $
For getting two roots of given equation
Taking \[ + \]sign on right side of equation $ \left( 2 \right) $
 $ \Rightarrow x - \dfrac{5}{6} = \dfrac{1}{6} $
Now we shift $ \dfrac{5}{6} $ on the right side of the equation then we get
 $ \Rightarrow x = \dfrac{1}{6} + \dfrac{5}{6} $
By taking 6 as LCM we get
 $
   \Rightarrow x = \dfrac{{1 + 5}}{6} \\
   \Rightarrow x = \dfrac{6}{6} \\
   \Rightarrow x = 1 \;
  $
Taking -sign on right side of equation $ \left( 2 \right) $
 $ \Rightarrow x - \dfrac{5}{6} = - \dfrac{1}{6} $
Now we shift $ \dfrac{5}{6} $ on the right side of the equation then we get
 $ \Rightarrow x = - \dfrac{1}{6} + \dfrac{5}{6} $
By taking 6 as LCM we get
 $
   \Rightarrow x = \dfrac{{ - 1 + 5}}{6} \\
   \Rightarrow x = \dfrac{4}{6} \\
   \Rightarrow x = \dfrac{2}{3} \;
  $
Hence $ x = 1 $ and $ x = \dfrac{2}{3} $ are the required roots of the given equation.
So, the correct answer is “$ x = 1 $ and $ x = \dfrac{2}{3} $”.

Note: Since In the given problem we have a polynomial of degree two therefore it is a quadratic equation. So the given equation has two roots because the number of roots of a polynomial is always the same as the degree of the polynomial.