Tag: approximation of square roots

Questions Related to approximation of square roots

Find the square root by guessing the units and tens digit:
$2116$
maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

  1. $46$

  2. $50$

  3. $52$

  4. $58$

Show answer
Correct Option: A
Explanation:

Its unit digit is 6 and tens digit is 1


For unit digit 6 , there are two possibilities either the square root end with 6 or 4

And according to options $46$ is the correct option

The number which exceeds its positive square root by $12$ is

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

  1. $9$

  2. $16$

  3. $25$

  4. None of these

Show answer
Correct Option: B
Explanation:
Let the positive number be x according to question,

$\sqrt{x}+12=x$

$\Rightarrow \sqrt{x}=x-12$

Squaring both sides,

$\Rightarrow x=x^{2}-24x+144$

$x^{2}-25x+144=0$

$x^{2}-16x-9x+144=0$

$x(x-16)-9(x-16)=0$

$(x-9)(x-16)=0$

 $  (x-9)=0 $ or $    (x-16)=0 $

 $ x=9 $  or $ x=16 $

So, the number is $16$.

Find the square root of which of the following numbers will be the least :

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

  1. $7\dfrac{58}{81}$

  2. $11\dfrac{14}{25}$

  3. $10\dfrac{1}{36}$

  4. $0.3481$

Show answer
Correct Option: D
Explanation:

$A.$


$7\dfrac{58}{81}=\dfrac{625}{81}$


$\Rightarrow$  $\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}=2.77$

$B.$

$11\dfrac{14}{25}=\dfrac{289}{25}$

$\Rightarrow$  $\sqrt{\dfrac{289}{25}}=\dfrac{17}{5}=3.4$

$C.$

$10\dfrac{1}{36}=\dfrac{361}{36}$

$\Rightarrow$  $\sqrt{\dfrac{361}{36}}=\dfrac{19}{6}=3.16$

$D.$

$0.3481=\dfrac{3481}{10000}$

$\Rightarrow$  $\sqrt{\dfrac{3481}{10000}}=\dfrac{59}{100}=0.59$

$\therefore$  We can see, $0.3481$  has least square root.

Simlify: $\sqrt{\dfrac{-17}{144}-i}$

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

  1. $ \pm \left( {\dfrac{3}{2} - \dfrac{i}{3}} \right)$

  2. $ \pm \left( {\dfrac{3}{4} - \dfrac{{2i}}{3}} \right)$

  3. $ \pm \left( {\dfrac{3}{5} - \dfrac{{5i}}{6}} \right)$

  4. $ \pm \left( {\dfrac{2}{3} - \dfrac{{3i}}{4}} \right)$

Show answer
Correct Option: D
Explanation:

We know that,

${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

 

Now, let,

$ -2ab=-i $

$ ab=i $

 

Now, consider $\dfrac{-17}{144}$. We can write it as,

$\dfrac{-17}{144}=\dfrac{64-81}{9\times 16}=\dfrac{4}{9}-\dfrac{9}{16}$

 

Thus,

$ \sqrt{\dfrac{-17}{144}-i}=\sqrt{\dfrac{4}{9}-\dfrac{9}{16}-i} $

$ =\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( i \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{2}}-2\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3i}{4} \right)} $

$ =\sqrt{{{\left( \dfrac{2}{3}-\dfrac{3i}{4} \right)}^{2}}} $

$ =\pm \left( \dfrac{2}{3}-\dfrac{3i}{4} \right) $

 

Hence, this is the required result.

The approximate value of$\sqrt { { \left( 1.97 \right)  }^{ 2 }{ \left( 4.02 \right)  }^{ 2 }{ \left( 3.98 \right)  }^{ 2 } }$

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

  1. $31.59 $

  2. $5.099$

  3. $5.009$

  4. $5.734$

Show answer
Correct Option: A
Explanation:

We have,

$ \sqrt{{{\left( 1.97 \right)}^{2}}{{\left( 4.02 \right)}^{2}}{{\left( 3.98 \right)}^{2}}} $

$ =1.97\times 4.02\times 3.98 $

$ =31.59 $

Hence, this is the answer.

Find the square root correct upto $2$ decimals $60.92$.

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

  1. $7.92$

  2. $7.81$

  3. $7.27$

  4. $7.56$

Show answer
Correct Option: B
Explanation:
$\sqrt{6092}$

$=\sqrt{\dfrac{6092}{100}}$

$=\dfrac{78.05}{10}$

$=7.81$

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

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

  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. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } + \sqrt { 3 } )$

Show answer
Correct Option: A
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 $A$

If $\sqrt{2}=1.414$ then the value of $\sqrt{8}$ is

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

  1. $2.828$

  2. $1.828$

  3. $2.282$

  4. $2.288$

Show answer
Correct Option: A
Explanation:
$\sqrt{2}=1.414$(given)
$\sqrt{8}=\sqrt{2\times 2\times 2}=2\sqrt{2}=2\times 1.414=2.828$
$\therefore \sqrt{8}=2.828$

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

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

  1. $\sqrt{13}-\sqrt{2}$

  2. $\sqrt{3}-\sqrt{2}$

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

  4. $\sqrt{5}-\sqrt{2}$

Show answer
Correct Option: B
Explanation:

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


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

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


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

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

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

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

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

  1. Always zero

  2. Never zero

  3. Positive if and only if a > b

  4. Positive only if a $\ne$ b

Show answer
Correct Option: D
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.