Question

Solve: $ {x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0 $

Answer 1

Hint:
This type of question can be solved in various ways and here we are going to solve the above equation by adding some number which is $ {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2} $ both the side and then solve it. And by using some simple formula we will get to the results.

Formula used:
The formula used in this question will be:
 $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $

Complete step by step solution:
 So we have an equation $ {x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0 $
So the above condition can likewise be composed as
 $ {x^2} - \left( {\sqrt 3 + 1} \right)x = - \sqrt 3 $
Now on adding $ {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2} $ both the sides of the equation, we get
\[ \Rightarrow {x^2} - 2\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)x + {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2} = - \sqrt 3 + {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2}\]
Presently as we probably aware of the formula that $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $
So on applying the formula in the above equation, we get
 $ \Rightarrow {\left( {x - \dfrac{{\sqrt 3 + 1}}{2}} \right)^2} = \dfrac{{ - 4\sqrt 3 + {{\left( {\sqrt 3 + 1} \right)}^2}}}{4} $
So the above condition can likewise be solved as and we get
\[ \Rightarrow {\left( {x - \dfrac{{\sqrt 3 + 1}}{2}} \right)^2} = {\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)^2}\]
Now on removing the squares from both sides, we get
\[ \Rightarrow \left( {x - \dfrac{{\sqrt 3 + 1}}{2}} \right) = \pm \left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)\]
Presently on taking the variable term one side, and illuminating for it we get
\[ \Rightarrow x = \dfrac{{\sqrt 3 + 1}}{2} \pm \dfrac{{\sqrt 3 - 1}}{2}\]
So now on solving for the value of $ x $ , we will get the two values of $ x $
We get
\[ \Rightarrow x = \sqrt 3 ,1\]

Hence, the roots of the equation $ {x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0 $ will be \[\sqrt 3 {\text{ and }}1\].

Note:
We have seen how easily we have solved it from this method but there is also another way through which we can solve such equations. It very well may be understood by utilizing the quadratic condition equation. The method which we had used above is completing the square method. And this method requires more calculations so if there is not compulsory to use the quadratic equation formula method then we should use the completing the square method to solve the problem and get through the solutions easily without any much more calculations.