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Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which of the following is smallest?

  1. $\sqrt [4]{5}$
  2. $\sqrt [5]{4}$
  3. $\sqrt {4}$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1 }{ 4 }  }\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 5 }  }\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 2 }  }\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1 }{ 2 }  }$

 
We now take the LCM of the denominators of the powers to make the denominators same, then the above magnitudes will be:

$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1\times 5 }{ 4\times 5 }  }={ \left( 5 \right)  }^{ \dfrac { 5 }{ 20 }  }={ \left( { 5 }^{ 5 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 3125 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 3125 } \\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 4 }{ 5\times 4 }  }={ \left( 4 \right)  }^{ \dfrac { 4 }{ 20 }  }={ \left( { 4 }^{ 4 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 256 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 256 } \\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 4 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 4 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 1048576 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 1048576 } \\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 3 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 3 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 2187 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 2187 }$     

Now, the descending order is as shown below:

$\sqrt [ 20 ]{ 1048576 } >\sqrt [ 20 ]{ 3125 } >\sqrt [ 20 ]{ 2187 } >\sqrt [ 20 ]{ 256 } \\ \Rightarrow \sqrt { 4 } >\sqrt [ 4 ]{ 5 } >\sqrt { 3 } >\sqrt [ 5 ]{ 4 }$ 

Hence, the smallest magnitude is $\sqrt [ 5 ]{ 4 }$.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Arrange the following surds in ascending order of their magnitudes: $\sqrt{5},\sqrt [ 3 ]{ 11 } ,2\sqrt [ 6 ]{ 3 } $

  1. $\sqrt [ 3 ]{ 11 } > \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  2. $\sqrt [ 3 ]{ 11 } < \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  3. $\sqrt [ 3 ]{ 11 } > \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
  4. $\sqrt [ 3 ]{ 11 } < \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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 $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Write $\displaystyle \sqrt[4]{6},\sqrt{2},\sqrt[3]{4}$ in ascending order

  1. $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$
  2. $\displaystyle \sqrt[4]{6}$, $\sqrt{2}$ and $\displaystyle \sqrt[3]{4}$
  3. $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$ and $\sqrt[4]{6}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
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 $\displaystyle \sqrt[3]{4}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is greater?
${ \left( \cfrac { 1 }{ 2 }  \right)  }^{ 1/2 } $ or ${ \left( \cfrac { 2 }{ 3 }  \right)  }^{ 1/3 } $

  1. ${ \left( \cfrac { 2 }{ 3 } \right) }^{ 1/3 } $
  2. ${ \left( \cfrac { 1 }{ 2 } \right) }^{ 1/2 } $

  3. Both are equal

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \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}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The correct descending order of the following surds is 

$ \sqrt [3]{2}$, $\sqrt 3$, $\sqrt 4$, $\sqrt 5$

  1. $\sqrt 3$ > $\sqrt 4$ > $\sqrt 5$ > $\sqrt [3]{2}$
  2. $\sqrt 4$ > $\sqrt 5$ > $\sqrt [3]{2}$ > $\sqrt 3$
  3. $\sqrt 5$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt [3]{2}$
  4. $\sqrt [3]{2}$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

$\therefore 125 > 64 > 27 > 4$

So, option $C$ is correct.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which one of the following is an irrational number?

  1. $\sqrt[3]{-27}$
  2. $\sqrt{2}(3\sqrt{2}+2\sqrt{8})$
  3. $\dfrac{3\sqrt{18}}{2\sqrt{6}}$
  4. $\sqrt{\dfrac{1}{2}}\cdot\sqrt{\dfrac{25}{2}}$
  5. $\dfrac{2\sqrt{5}}{\sqrt{45}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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 $C$.

Multiple choice maths factorization-1 solution of a quadratic equation by completing the square solving a quadratic equation solving a quadratic equation: completing the square

Solve the following quadratic equation by completing the square: $ x^2+(\sqrt{3}+1)x+\sqrt{3}=0$

  1. $\left \{ \sqrt{3}, 1\right \}$
  2. $\left \{ \sqrt{3}, 2\right \}$
  3. $\left \{ \sqrt{3}, 3\right \}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


$\Rightarrow$  $x^2-(\sqrt{3}+1)x=-\sqrt{3}$


Now, adding $\left(\dfrac{\sqrt{3}+1}{2}\right)^2$ on both sides,

$\Rightarrow$  $x^2-2\times x\times \left(\dfrac{\sqrt{3}+1}{2}\right)+\left(\dfrac{\sqrt{3}+1}{2}\right)^2=-\sqrt{3}+\left(\dfrac{\sqrt{3}+1}{2}\right)^2$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{(\sqrt{3}+1)^2}{4}-\sqrt{3}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3-2\sqrt{3}+1}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{\sqrt{3}-1}{2}\right]^2$

Taking square root on both sides, 
  
$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\pm\dfrac{\sqrt{3}-1}{2}$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\dfrac{\sqrt{3}-1}{2}$ 
and $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=-\dfrac{\sqrt{3}-1}{2}$

$\therefore$  $x=\dfrac{2\sqrt{3}}{2}$  and $ x=\dfrac{2}{2}$

$\therefore$  $x=\sqrt{3}$ and $x=1$