Question

Find the roots of $ 4{x^2} + 3x + 5 = 0 $ by the method of completing the square.

Answer 1

Hint: In completing square method we first make leading coefficient of an equation one, if it is not one by dividing whole equation with leading coefficient and then adding square of half of coefficient of ‘x’ on both side to get square on left hand side and then solving further by taking square root to get roots of given equation.

Complete step-by-step answer:
Given quadratic equation is $ 4{x^2} + 3x + 5 = 0 $
There are different steps to find roots of a given quadratic equation by completing the square method.
In this method we first make the shift constant term to the right hand side if it is on left side.
 $ 4{x^2} + 3x = - 5 $
In the second step we remove the coefficient of $ {x^2} $ if it is there or if not then proceed to the third step.
As a given problem coefficient of $ {x^2}\,\,is\,\,4. $
So, we remove it or either we make it one.
Divide each term by $ 4 $ to make the coefficient of $ {x^2}\,as\,\,1. $
 $
   \Rightarrow \dfrac{{4{x^2}}}{4} + \dfrac{{3x}}{4} = - \dfrac{5}{4} \\
  or \\
  {x^2} + \dfrac{3}{4}x = - \dfrac{5}{4} \;
  $
In the third step we add a square of half of coefficient of ‘x’ on both side of above equation.
 $
   \Rightarrow {x^2} + \dfrac{3}{4}x + {\left[ {\dfrac{1}{2}\left( {\dfrac{3}{4}} \right)} \right]^2} = - \dfrac{5}{4} + {\left[ {\dfrac{1}{2}\left( {\dfrac{3}{4}} \right)} \right]^2} \\
   \Rightarrow {x^2} + \dfrac{3}{4}x + {\left( {\dfrac{3}{8}} \right)^2} = - \dfrac{5}{4} + \dfrac{9}{{64}} \\
   \Rightarrow {\left( {x + \dfrac{3}{8}} \right)^2} = \dfrac{{ - 80 + 9}}{{64}} \\
   \Rightarrow {\left( {x + \dfrac{3}{8}} \right)^2} = - \dfrac{{71}}{{64}} \;
  $
Clearly in the above equation we see that the left hand side is square and the right hand side is a negative. Which can’t be possible.
Hence, from above we can say that the roots of a given quadratic equation is not real.
Roots of quadratic equation $ 4{x^2} + 3x + 5 = 0 $ are complex roots.
These complex roots can be found by taking the square root of the right hand side.
 $
   \Rightarrow x + \dfrac{3}{8} = \sqrt { - \dfrac{{71}}{{64}}} \\
   \Rightarrow x + \dfrac{3}{8} = \pm \dfrac{{\sqrt {71} i}}{8} \;
  $
 $
   \Rightarrow x + \dfrac{3}{8} = \dfrac{{\sqrt {71} i}}{8}\,\,\,\,or\,\,\,\,x + \dfrac{3}{8} = \dfrac{{ - \sqrt {71} i}}{8} \\
   \Rightarrow x = - \dfrac{3}{8} + \dfrac{{\sqrt {71} i}}{8}\,\,or\,\,x = - \dfrac{3}{8} - \dfrac{{\sqrt {71} i}}{8} \\
   \Rightarrow x = \dfrac{{ - 3 + \sqrt {71} i}}{8}\,\,or\,\,x = \dfrac{{ - 3 - \sqrt {71} i}}{8} \\
  $
Therefore, from above we can say that roots of the given quadratic equation are $ \dfrac{{ - 3 + \sqrt {71} i}}{8}\,\,and\,\,\dfrac{{ - 3 - \sqrt {71} i}}{8} $ .
So, the correct answer is “$ \dfrac{{ - 3 + \sqrt {71} i}}{8}\,\,and\,\,\dfrac{{ - 3 - \sqrt {71} i}}{8} $ ”.

Note: There are different methods to find zeros or roots of a given quadratic equation like the middle term splitting method. In this we write the middle term as the sum or difference of two numbers such that the resulting equation can be written in factored form and hence we can get zeros or roots. In quadratic formula method we first find discriminant of given quadratic equation and then using values in $ \dfrac{{ - b \pm \sqrt D }}{{2a}} $ to get zeros or roots of equation and third method is completing square method whose explanation is given above.