Tag: squares and square roots

Quizzes (7) Questions (116)

Questions Related to squares and square roots

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The positive square root of $( \sqrt { 48 } - \sqrt { 45 } )$ is _________.

  1. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } - \sqrt { 3 } )$
  2. $\frac { \sqrt [ 4 ] { 3 } } { 2 } ( \sqrt { 5 } - \sqrt { 3 } )$
  3. $\frac { \sqrt { 2 } } { \sqrt [ 4 ] { 3 } } ( \sqrt { 5 } - \sqrt { 3 } )$
  4. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } + \sqrt { 3 } )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving $(\sqrt{48}-\sqrt{45})$

$4\sqrt{3}-3\sqrt{5}$

$\dfrac{\sqrt{3}}{2}(8-2\sqrt{15})$

$\dfrac{\sqrt{3}}{2}(3+5-2\sqrt{15})$

$\dfrac{\sqrt{3}}{2}(\sqrt{3^2}+\sqrt{5^2}-2\sqrt{5}\times\sqrt{3})$

$\dfrac{\sqrt3}{2}(\sqrt5-\sqrt3)^2$

$Now \ Finding\ Square \ Root$

$\pm{ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)}$

$So \ it's \ positive \ root \ is \ $$ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)$
Correct Answer is $A$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Find the square root of 
$5-2\sqrt{6}$

  1. $\sqrt{13}-\sqrt{2}$
  2. $\sqrt{3}-\sqrt{2}$
  3. $\sqrt{5}-\sqrt{3}$
  4. $\sqrt{5}-\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$5-2\sqrt{6}=3+2-2\sqrt{6}$


$=({\sqrt{3}})^2+({\sqrt{2}})^2-2\sqrt{3}\times \sqrt {2}$

Using $(a-b)^2=a^2+b^2-2ab$


${5-2\sqrt{6}}=(\sqrt{3}-\sqrt{2})^2$

$\sqrt {5-2\sqrt{6}}=(\sqrt{3}-\sqrt{2})$

$So, \  the \  square \  root \  of \  (5-2\sqrt{6})=\sqrt{3}-\sqrt{2}$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$ is

  1. Always zero

  2. Never zero

  3. Positive if and only if a > b
  4. Positive only if a $\ne$ b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$
$= |a - b| + |b - a|$

Now, If $a > b$
$= a - b + a - b$
$= 2a - 2b$...+ ve

If $b > a$
$= b - a + b - a$
$= 2b - 2a$...+ ve
Therefore, if $a \ne b$ then the given equation is always positive.
Hence, option 'D' is correct.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

If $\sqrt{6}\, =\, 2.55,$ then the value of $\displaystyle {\sqrt{\frac{2}{3}\, +\, 3\frac{3}{2}}}$ is

  1. 4.48

  2. 4.49

  3. 4.50

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$ {\sqrt{\cfrac{2}{3} + 3\cfrac{3}{2}}}$
$=  {\cfrac{\sqrt{2}}{\sqrt{3}} \times \cfrac{\sqrt{3}}{\sqrt{3}} + 3 \times \cfrac{\sqrt{3}}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}}}$
$=  {\cfrac{\sqrt{6}}{3} + \cfrac{3\sqrt{6}}{2} = \cfrac{2.55}{3} + \cfrac{3 \times 2.55}{2}}$
$=  {\cfrac{2.55}{3} + \cfrac{7.65}{2} = \cfrac{5.10 + 22.95}{6}}$
$=  \cfrac{28.05}{6} = 4.675$
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\displaystyle {\sqrt{\frac{4}{3}}\, -\, \sqrt{\frac{3}{4}}\, =\, ?}$

  1. $\displaystyle \frac{1}{2\sqrt{3}}$
  2. $\displaystyle - \frac{1}{2\sqrt{3}}$

  3. 1
  4. $\displaystyle \frac{5\sqrt{3}}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle {\frac{\sqrt{4}}{\sqrt{3}} - \frac{\sqrt{3}}{\sqrt{4}} = \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{4 - 3}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}}$