
Arrange in ascending order $\sqrt[3]{2},\sqrt 3 ,\sqrt[6]{5}$ .
Answer
474.9k+ views
Hint: In this question we have been given the number under the roots. So we will first write the numbers in the simpler form and then we will try to make the denominators of the exponents equal. After that, we will check the values which are smaller and then we will write them in ascending order.
Complete step-by-step answer:
Here we have : $\sqrt[3]{2},\sqrt 3 ,\sqrt[6]{5}$ .
We can also write the expression in simpler form as follows:
${2^{\dfrac{1}{3}}},{3^{\dfrac{1}{2}}},{5^{\dfrac{1}{6}}}$
Now we can see that in the denominators of the exponents we have: $3,2,6$
We will now take the LCM of $3,2,6$ , which is $6$ .
Now we will make all the denominators equal to $6$ , so we have to multiply by the multiples in both numerator and denominator.
We can write the numbers as:
${2^{\dfrac{1}{3} \times \dfrac{2}{2}}} = {2^{\dfrac{2}{6}}}$
For the second number we can write:
${3^{\dfrac{1}{2} \times \dfrac{3}{3}}} = {3^{\dfrac{3}{6}}}$
Since in the third number we already have the desired denominator, so the third number is
${5^{\dfrac{1}{6}}}$
Now we will again write the numbers in the root under, but we have to keep in mind that the numerator will turn as the exponential powers inside the root.
So we have the numbers as:
$\sqrt[6]{{{2^2}}},\sqrt[6]{{{3^3}}},\sqrt[6]{5}$
We will simplify the values inside the root, so we have:
$\sqrt[6]{4},\sqrt[6]{{27}},\sqrt[6]{5}$
From this we can write the smaller value in the front and then the larger value:
$\sqrt[6]{4},\sqrt[6]{5},\sqrt[6]{{27}}$
Hence the original numbers in ascending form are:
$\sqrt[3]{2},\sqrt[6]{5},\sqrt 3 $ .
Note: We should note that The symbol $\sqrt {} $ is used to denote square root and this symbol of square roots is also known as radical. The number written inside the square root symbol or radical is known as radicand. We know that all real numbers have two square roots, one is a positive square root and another one is a negative square root. The positive square root is also referred to as the principal square root.
Complete step-by-step answer:
Here we have : $\sqrt[3]{2},\sqrt 3 ,\sqrt[6]{5}$ .
We can also write the expression in simpler form as follows:
${2^{\dfrac{1}{3}}},{3^{\dfrac{1}{2}}},{5^{\dfrac{1}{6}}}$
Now we can see that in the denominators of the exponents we have: $3,2,6$
We will now take the LCM of $3,2,6$ , which is $6$ .
Now we will make all the denominators equal to $6$ , so we have to multiply by the multiples in both numerator and denominator.
We can write the numbers as:
${2^{\dfrac{1}{3} \times \dfrac{2}{2}}} = {2^{\dfrac{2}{6}}}$
For the second number we can write:
${3^{\dfrac{1}{2} \times \dfrac{3}{3}}} = {3^{\dfrac{3}{6}}}$
Since in the third number we already have the desired denominator, so the third number is
${5^{\dfrac{1}{6}}}$
Now we will again write the numbers in the root under, but we have to keep in mind that the numerator will turn as the exponential powers inside the root.
So we have the numbers as:
$\sqrt[6]{{{2^2}}},\sqrt[6]{{{3^3}}},\sqrt[6]{5}$
We will simplify the values inside the root, so we have:
$\sqrt[6]{4},\sqrt[6]{{27}},\sqrt[6]{5}$
From this we can write the smaller value in the front and then the larger value:
$\sqrt[6]{4},\sqrt[6]{5},\sqrt[6]{{27}}$
Hence the original numbers in ascending form are:
$\sqrt[3]{2},\sqrt[6]{5},\sqrt 3 $ .
Note: We should note that The symbol $\sqrt {} $ is used to denote square root and this symbol of square roots is also known as radical. The number written inside the square root symbol or radical is known as radicand. We know that all real numbers have two square roots, one is a positive square root and another one is a negative square root. The positive square root is also referred to as the principal square root.
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