Tag: surds and law of surds

Questions Related to surds and law of surds

Which of the following is smallest?

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

  1. $ \displaystyle \sqrt[4]{5} $

  2. $ \displaystyle \sqrt[5]{4} $

  3. $ \displaystyle \sqrt{4} $

  4. $ \displaystyle \sqrt{3} $

Show answer
Correct Option: B
Explanation:

Since, fifth root of four is smallest.

Option $B$ is correct.

Which one of the following is the smallest surd?

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

  1. $\sqrt 4$

  2. $\sqrt {27}$

  3. $\sqrt 9$

  4. $\sqrt 5$

Show answer
Correct Option: A
Explanation:
As we know, $4 < 5 < 9 < 27$
So, $\sqrt 4$ < $\sqrt 5$ < $\sqrt 9$ < $\sqrt {27}$
Hence, the answer is $\sqrt{4}$.

If $p=\sqrt{32}-\sqrt{24}$ and $q=\sqrt{50}-\sqrt{48}$, then:

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

  1. $p< q$

  2. $p> q$

  3. $p=q$

  4. $p\le q$

Show answer
Correct Option: B
Explanation:

$p= \sqrt{32} - \sqrt{24}$

$p= 5.656-4.898 = 0.758$

$q = \sqrt{50} + \sqrt{48}$

$q= 7.071- 6.928 = 0.089$

$\therefore p> q$

If $x=\sqrt{2}+1, y=\sqrt{17}-\sqrt{2}$, then .............

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

  1. $x< y$

  2. $x>y$

  3. $x=y$

  4. $x\ge y$

Show answer
Correct Option: A
Explanation:

$\sqrt{2} =1.414$

$\sqrt{17} =4.123$

Hence,

$x = \sqrt{2}+1 =1.414+1 = 2.414$

$ y= \sqrt{17} -\sqrt{2} =4.123- 1.414 = 2.709$

Hence , $x<y$

$\sqrt{11}-\sqrt{10}$ $\Box$ $ \sqrt{12}-\sqrt{11}$

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

  1. $<$

  2. $>$

  3. $=$

  4. cannot be determined

Show answer
Correct Option: B
Explanation:

$\sqrt{11} =3.316$

$\sqrt{10} =3.162$

$\sqrt{12} =3.464$

Hence,

$\sqrt{11} -\sqrt{10} =3.316-3.162 = 0.154$

$\sqrt{12} -\sqrt{11} =3.464-3.316 = 0.148$

Hence , $\sqrt{11} -\sqrt{10} >\sqrt{12} -\sqrt{11}$

Which among the following numbers is the greatest?
$\displaystyle \sqrt[3]{4},\sqrt{2},\sqrt[6]{13},\sqrt[4]{5}$

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

  1. $\displaystyle \sqrt[3]{4}$ is the greatest

  2. $\sqrt{2}$ is the greatest

  3. $\sqrt[6]{13}$ is the greatest

  4. $\sqrt[4]{5}$ is the greatest

Show answer
Correct Option: A
Explanation:

LCM of $3, 6, 4 = 12$

So, raising each given number to power $12$.
$\Rightarrow \sqrt[3]{4}=(4)^{1/3}=(4^{1/3})^{12}=4^{4}=256$
$\Rightarrow \sqrt{2}=(2)^{1/2}=(2^{1/2})^{12}=2^{6}=64$
$\Rightarrow \sqrt[6]{13}=(13)^{1/6}=(13^{1/6})^{12}=13^{2}=169$
$\Rightarrow \sqrt[4]{5}=(5)^{1/4}=(15^{1/4})^{12}=5^{3}=125$

$\therefore \sqrt[3]{4}$ is the greatest.

If $x=\sqrt{7}-\sqrt{5}, y=\sqrt{13}-\sqrt{11}$, then:

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

  1. $x=y$

  2. $x> y$

  3. $x< y$

  4. $x\ge y$

Show answer
Correct Option: B
Explanation:

$x = \sqrt{7}-\sqrt{5}$

$= 2.64 – 2.23= 0.41$

$y=  \sqrt{13}-\sqrt{11}$

$= 3.60- 3.31 = 0.29$

$Hence, x> y$

If $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}-\sqrt{5}$, then identify the true statement.

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

  1. $A> B$

  2. $A=B$

  3. $A< B$

  4. $A\ge B$

Show answer
Correct Option: C
Explanation:
$A=\sqrt{7}-\sqrt{6}\Rightarrow \dfrac{1}{A}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$
$\Rightarrow \boxed{\dfrac{1}{A}=\sqrt{7}+\sqrt{6}}$
$\boxed{\dfrac{1}{B}=\sqrt{6}+\sqrt{5}}$
As $\sqrt{7} > \sqrt{5}\Rightarrow \dfrac{1}{A} > \dfrac{1}{B}$
$\Rightarrow \boxed{B > A}$

The smallest between $\sqrt{17} - \sqrt{12}$ and $\sqrt{11} - \sqrt{6}$ is _________.

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

  1. $\sqrt{17} - \sqrt{12}$

  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined

Show answer
Correct Option: A
Explanation:

$\sqrt{17} \approx 4.123$

$\sqrt{12} \approx 3.464$
$\sqrt{11} \approx 3.316$
$\sqrt{6} \approx 2.449$

$\Rightarrow \sqrt{17} - \sqrt{12} = 0.659$
$\Rightarrow \sqrt{11} - \sqrt{6} = 0.867$

Hence, $\sqrt{17}-\sqrt{12}$ is smaller.

The smallest of $\sqrt [ 3 ]{ 4 } , \sqrt [ 4 ]{ 5 } , \sqrt [ 4 ]{ 6 } , \sqrt [ 3 ]{ 8 } $ is:

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

  1. $\sqrt [ 3 ]{ 8 } $

  2. $\sqrt [ 4 ]{ 5 } $

  3. $\sqrt [ 3 ]{ 4 } $

  4. $\sqrt [ 4 ]{ 6 } $

Show answer
Correct Option: B
Explanation:
$\sqrt[3]{4}=\sqrt[12]{44}=\sqrt[12]{256}$
$\sqrt[4]{6}=\sqrt[12]{5^3}=\sqrt[12]{125}$
$\sqrt[4]{6}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$
$\sqrt[3]{8}=\sqrt[12]{8^{4}}=\sqrt[12]{64^{2}}$
As $'125'$ is smallest
$\therefore \boxed{4\sqrt{5}}$ is smallest