Tag: maths

Let us rewrite the given set of magnitudes $\sqrt [ 4 ]{ 5 } ,\sqrt [ 5 ]{ 4 } ,\sqrt { 4 },\sqrt {3}$ as follows:

$\sqrt{5} = 5^{1/2}$

$\sqrt[3]{11} = 11^{1/3}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5} = 5^{\frac{1}{2}\times\frac{6}{6}} = \sqrt [12]{5^6}=\sqrt[12]{15625}$

$\sqrt[3]{11} = 11^{\frac{1}{3}\times\frac{4}{4}} = \sqrt [12]{11^4} =\sqrt[12]{14641}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{\frac{1}{6}\times\frac{2}{2}} = \sqrt[12]{12^2} =\sqrt[12]{144}$

Hence, the Ascending order is $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
The surds are of the order $4, 2$ and $3$ respectively The L.C.M. is $12$ So, we change each surd of the order $12$
The terms now are $\displaystyle \left ( 6^{3} \right )^\cfrac {1}{12},\left ( 2^{6} \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 4^{4} \right )^\cfrac{1}{12}$
$\displaystyle \Rightarrow \left ( 216 \right )^\cfrac{1}{12},\left ( 64 \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 256 \right )^\cfrac{1}{12}$
$\displaystyle \therefore $ Ascending order is $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$

$ \left(\dfrac{1}{2}\right)^{1/2} \; or \; \left(\dfrac{2}{3}\right)^{1/3}$

$= \left(\left(\dfrac{1}{2}\right)^{1/2}\right)^6 \; or \; \left(\left(\dfrac{2}{3}\right)^{1/3}\right)^6$

$= \left(\dfrac{1}{2}\right)^3 \; or \; \left(\dfrac{2}{3}\right)^2$

$= \left(\dfrac{1}{8}\right) \; or \; \left(\dfrac{4}{9}\right)$

= 0.125 or 0.44

Since, 0.44 is greater and so is $ \left(\dfrac{2}{3}\right)^{1/3}$

LCM of $3$ and $2$ is $6$.
Therefore, multiplying the index of all numbers by $6$.
${\sqrt [3]{2}}^6 = 2^\dfrac 63 = 2^2 = 4$

${\sqrt 3}^6 = 3^\dfrac 62 = 3^3 = 27$

${\sqrt 4}^6 = 4^ \dfrac 62 = 4^3 = 64$

${\sqrt 5}^6 = 5^{\dfrac 62} =  5^3 = 125$

$\therefore 125 > 64 > 27 > 4$

Option A: $\sqrt [ 3 ]{ -27 } ={ (-3) }^{ 3\times \frac { 1 }{ 3 }  }=-3$
Option B: $\sqrt { 2 } (3\sqrt { 2 } +2\sqrt { 8 } )=\sqrt { 2 } (3\sqrt { 2 } +4\sqrt { 2 } )=\sqrt { 2 } (7\sqrt { 2 } )=14$
Option C: $\dfrac { 3\sqrt { 18 }  }{ 2\sqrt { 6 }  } =\dfrac { 3\sqrt { 3 }  }{ 2 } $
Option D: $\sqrt { \dfrac { 1 }{ 2 }  } \sqrt { \dfrac { 25 }{ 2 }  } =\dfrac { 5 }{ 2 } $
Option E: $\dfrac { 2\sqrt { 5 }  }{ \sqrt { 45 }  } =\dfrac { 2\sqrt { 5 }  }{ 3\sqrt { 5 }  } =\dfrac { 2 }{ 3 } $
Therefore, all are rational except option $C$.

Given equation is $x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

Zero degree polynomial is considered as constant.
Example: $p(x)= k$
Therefore, $k$ is constant.
The constant term in an expression or equation has a fixed value and does not contain variables.