
How do you rationalize the denominator $\dfrac{3}{{\sqrt 3 }}?$
Answer
488.1k+ views
Hint:To rationalize the denominator of the given fraction, first find the conjugate part of the denominator of the given fraction. Then multiply and divide the fraction with the conjugate part.Also conjugate part of the irrational number $a \pm \sqrt b $ is given as $a \mp \sqrt b $.
Complete step by step answer:
In order to rationalize the denominator and simplify the given expression $\dfrac{3}{{\sqrt 3 }}$, we will first find the conjugate part of the denominator, that is $\sqrt 3 $.Conjugate part of $\sqrt 3 $ is equals to $\sqrt 3 $.Now, multiplying and dividing the given expression with conjugate part, in order to rationalize the denominator
$\dfrac{3}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} \\
\Rightarrow\dfrac{{3\sqrt 3 }}{3} \\ $
Now, taking out three common from the numerator part and the denominator part in order to simplify the fraction further, we will get
$\dfrac{{3\sqrt 3 }}{3} \\
\therefore\sqrt 3 \\ $
Therefore $\sqrt 3 $ is the rationalized form of $\dfrac{3}{{\sqrt 3 }}$.
Note:You may be thinking of why we have taken the conjugate part of $\sqrt 3 $ is equals to $\sqrt 3 $ even when we have described above that conjugate part of $a + \sqrt b $ is given as $a - \sqrt b $. This is why because in this problem value of a is zero and the conjugate part will become $ - \sqrt b $ and when we multiply and divide the conjugate part, the negative sign will get cancel out from itself, that’s why we have taken directly $\sqrt 3 $ as the conjugate part of $\sqrt 3 $. You can also take conjugate part like this but only condition with this is the value of “a” should equal to zero. This problem can be directly rationalized by expressing the numerator as a product of the denominator, and then cancelling out $\sqrt 3 $ of the numerator with the denominator.
Complete step by step answer:
In order to rationalize the denominator and simplify the given expression $\dfrac{3}{{\sqrt 3 }}$, we will first find the conjugate part of the denominator, that is $\sqrt 3 $.Conjugate part of $\sqrt 3 $ is equals to $\sqrt 3 $.Now, multiplying and dividing the given expression with conjugate part, in order to rationalize the denominator
$\dfrac{3}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} \\
\Rightarrow\dfrac{{3\sqrt 3 }}{3} \\ $
Now, taking out three common from the numerator part and the denominator part in order to simplify the fraction further, we will get
$\dfrac{{3\sqrt 3 }}{3} \\
\therefore\sqrt 3 \\ $
Therefore $\sqrt 3 $ is the rationalized form of $\dfrac{3}{{\sqrt 3 }}$.
Note:You may be thinking of why we have taken the conjugate part of $\sqrt 3 $ is equals to $\sqrt 3 $ even when we have described above that conjugate part of $a + \sqrt b $ is given as $a - \sqrt b $. This is why because in this problem value of a is zero and the conjugate part will become $ - \sqrt b $ and when we multiply and divide the conjugate part, the negative sign will get cancel out from itself, that’s why we have taken directly $\sqrt 3 $ as the conjugate part of $\sqrt 3 $. You can also take conjugate part like this but only condition with this is the value of “a” should equal to zero. This problem can be directly rationalized by expressing the numerator as a product of the denominator, and then cancelling out $\sqrt 3 $ of the numerator with the denominator.
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