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

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$
  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

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

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

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

Compare the following pairs of surds $\sqrt[8]{12}, \sqrt[4]{6}$

  1. $\sqrt[8]{2} < \sqrt[4]{6}$
  2. $\sqrt[8]{8} < \sqrt[4]{6}$
  3. $\sqrt[8]{12} < \sqrt[4]{6}$
  4. $\sqrt[8]{12} < \sqrt[4]{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To compare, convert to a common root: sqrt[8]{12} is 12^(1/8) and sqrt[4]{6} is 6^(1/4) = 6^(2/8) = 36^(1/8). Since 12 < 36, sqrt[8]{12} < sqrt[4]{6}.

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

Compare the following pair of surds:

$\sqrt[3]{6}, \sqrt[4]{8}$

  1. $\sqrt[3]{6} > \sqrt[4]{8}$
  2. $\sqrt[3]{6} > \sqrt[4]{4}$
  3. $\sqrt[3]{6} > \sqrt[3]{8}$
  4. $\sqrt[3]{4} > \sqrt[4]{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to a common root (12th root): sqrt[3]{6} = 6^(4/12) = 1296^(1/12). sqrt[4]{8} = 8^(3/12) = 512^(1/12). Since 1296 > 512, the first is larger.

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

Arrange the following in ascending order of magnitude: 

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

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

Convert to 12th roots: sqrt[3]{4} = 4^(4/12) = 256^(1/12). sqrt[4]{5} = 5^(3/12) = 125^(1/12). sqrt{3} = 3^(6/12) = 729^(1/12). Ordering 125 < 256 < 729 gives the correct sequence.

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

What is the least value of $a$ in $ \displaystyle\frac{\sqrt 2+\sqrt 3}{\sqrt{2+3}} < a$?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2+3 }  } =\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 5 }  } =\dfrac { (\sqrt { 2 } +\sqrt { 3 } )\times \sqrt { 5 }  }{ 5 } =\dfrac { 7.02 }{ 5 } \\ =1.40$
$\Rightarrow 1.40<a$
So, least integer value of $a$ is $2$.
Hence, option B is correct.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The greatest among $\displaystyle \sqrt[6]{3}$, $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$, $\displaystyle \sqrt[4]{5}$ is--

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

$\displaystyle \therefore $ $\displaystyle \sqrt[6]{3}$ = $\displaystyle \left ( 3 \right )^{\dfrac{1}{6}}$ = $\displaystyle \left ( 3^{2} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 9 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt{2}$ = $\displaystyle \left ( 2 \right )^{\dfrac{1}{2}}$ = $\displaystyle \left ( 2^{6} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 64 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt[3]{4}$ = $\displaystyle \left ( 4 \right )^{\dfrac{1}{3}}$ = $\displaystyle \left ( 4^{4} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$

$\sqrt[4]{5}=(5)^{\dfrac{1}{4}}=(5^{3})^{\dfrac{1}{12}}=(125)^{\dfrac{1}{12}}$
$\displaystyle \therefore $ The greatest number is $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$ = $\displaystyle \sqrt[3]{4}$