
Find the solutions of $ 3{{x}^{2}}-2\sqrt{6}x+2=0 $ by the method of completing the squares when $ x $ is a real number.
(a) $ x $ = $ -\sqrt{\dfrac{2}{3}} $
(b) $ x $ = $ \pm \sqrt{\dfrac{2}{3}} $
(c) Cannot be determined
(d) None of these
Answer
487.2k+ views
Hint: To do this question we need to first learn about the method of completing the squares. In this method we first make the coefficient of $ {{x}^{2}} $ as 1 then we will take the constant part of the quadratic equation to the RHS side of the equation then we add or subtract constant number on both sides of the equation to make a perfect square on the LHS of the equation. When a perfect square is made on the LHS of the equation we will equate the square with whatever constant we have on our RHS and we will get the desired value of the variable.
Complete step-by-step answer:
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}} $
So, the correct answer is “Option D”.
Note: To solve by method of completing the squares you need to follow every step if you miss out any of the given steps you will not be able to solve the problems. Some students miss out even the first step i.e. to make the coefficient of the $ {{x}^{2}} $ as 1 and end up wasting a lot of time so try to go step by step only.
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.
Complete step-by-step answer:
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}} $
So, the correct answer is “Option D”.
Note: To solve by method of completing the squares you need to follow every step if you miss out any of the given steps you will not be able to solve the problems. Some students miss out even the first step i.e. to make the coefficient of the $ {{x}^{2}} $ as 1 and end up wasting a lot of time so try to go step by step only.
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.
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