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(a) $ x $ = $ -\sqrt{\dfrac{2}{3}} $

(b) $ x $ = $ \pm \sqrt{\dfrac{2}{3}} $

(c) Cannot be determined

(d) None of these

Answer

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We need solve the problem step by step, so

We will apply the method of completing the squares to the given question, we are given that

$ 3{{x}^{2}}-2\sqrt{6}x+2=0 $

First we need to make the constant of $ {{x}^{2}} $ as 1, so divide both sides by it’s coefficient i.e. 3, we get

\[{{x}^{2}}-\dfrac{2\sqrt{6}x}{3}+\dfrac{2}{3}=0\]

Taking constant part i.e. $ \dfrac{2}{3} $ to RHS, we get

\[{{x}^{2}}-\dfrac{2\sqrt{6}x}{3}=-\dfrac{2}{3}\]

Now adding square of $ -\dfrac{\sqrt{6}}{3} $ on both sides we get,

\[{{x}^{2}}-\dfrac{2\sqrt{6}x}{3}+{{\left( -\dfrac{\sqrt{6}}{3} \right)}^{2}}=-\dfrac{2}{3}+{{\left( -\dfrac{\sqrt{6}}{3} \right)}^{2}}\]

After further solving the above expression we get,

$ {{\left( x-\dfrac{\sqrt{6}}{3} \right)}^{2}}=0 $

Hence, value of $ x $ will be,

$ x=\dfrac{\sqrt{6}}{3}=\sqrt{\dfrac{2}{3}} $

You can also do this question by checking each option and if the option satisfies the equation then it will be our required answer but do by this method only in objective type because it is mentioned in the question itself that we have to do it by method of completing squares so in subjective follow the given method.