Question

Find the roots of the equation \[5{x^2} - 6x - 2 = 0\] by method of completing the square.
A) \[\dfrac{{2 + \sqrt 7 }}{5}\] and \[\dfrac{{3 + \sqrt {19} }}{5}\]
B) \[\dfrac{{3 + \sqrt {19} }}{5}\]and \[\dfrac{{3 - \sqrt {19} }}{5}\]
C) \[\dfrac{{3 - \sqrt {19} }}{5}\] and \[\dfrac{{8 - \sqrt {19} }}{5}\]
D) None of these

Answer 1

Hint: Roots of an equation are such values of the variable of equation which when substituted in the equation gives zero.

Complete step-by-step answer:
The given equation is:-
\[5{x^2} - 6x - 2 = 0\]
Let us first make the coefficient of \[{x^2}\] as 1.
Therefore dividing whole equation by 5 we get:-
\[
  \dfrac{{5{x^2} - 6x - 2}}{5} = \dfrac{0}{5} \\
   \Rightarrow \dfrac{5}{5}{x^2} - \dfrac{6}{5}x - \dfrac{2}{5} = 0 \\
   \Rightarrow {x^2} - \dfrac{6}{5}x - \dfrac{2}{5} = 0 \\
 \]
Now we will move the number term to the right side of the equation:-
\[ \Rightarrow {x^2} - \dfrac{6}{5}x = \dfrac{2}{5}\]
Now we have to make the square on the left side of the equation and balance this by adding the same number to the right side of the equation.
Hence adding and subtracting \[\dfrac{9}{{25}}\] on left hand side we get:-
\[ \Rightarrow {x^2} - \dfrac{6}{5}x + \dfrac{9}{{25}} - \dfrac{9}{{25}} = \dfrac{2}{5}\]
\[ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{3}{5}} \right)^2} - 2\left( x \right)\left( {\dfrac{3}{5}} \right) - \dfrac{9}{{25}} = \dfrac{2}{5}\]
Now as we know that:-
\[{a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}\]
Hence applying this identity in the above equation we get:-
\[ \Rightarrow {\left( {x - \dfrac{3}{5}} \right)^2} - \dfrac{9}{{25}} = \dfrac{2}{5}\]
Solving it further we get:-
\[{\left( {x - \dfrac{3}{5}} \right)^2} = \dfrac{9}{{25}} + \dfrac{2}{5}\]
Taking LCM we get:-
\[
  {\left( {x - \dfrac{3}{5}} \right)^2} = \dfrac{{9 + 2\left( 5 \right)}}{{25}} \\
   \Rightarrow {\left( {x - \dfrac{3}{5}} \right)^2} = \dfrac{{9 + 10}}{{25}} \\
 \]
Simplifying it further we get:-
\[{\left( {x - \dfrac{3}{5}} \right)^2} = \dfrac{{19}}{{25}}\]
Now taking square root of both the sides we get:-
\[
  \sqrt {{{\left( {x - \dfrac{3}{5}} \right)}^2}} = \sqrt {\dfrac{{19}}{{25}}} \\
   \Rightarrow \left( {x - \dfrac{3}{5}} \right) = \pm \dfrac{{\sqrt {19} }}{5} \\
 \]
Now solving for x we get:-
\[
  x = \dfrac{3}{5} \pm \dfrac{{\sqrt {19} }}{5} \\
   \Rightarrow x = \dfrac{{3 + \sqrt {19} }}{5};x = \dfrac{{3 - \sqrt {19} }}{5} \\
 \]
Hence the roots of the given equation are:-
\[x = \dfrac{{3 + \sqrt {19} }}{5};x = \dfrac{{3 - \sqrt {19} }}{5}\]

So, the correct answer is “Option B”.

Note: While taking the square root students may consider only the positive value that may lead to the wrong solution.
Whenever we take the square root of any quantity then both the positive and the negative values are to be considered.
\[\sqrt {{a^2}} = \pm a\]