Square Roots and Perfect Squares

$\sqrt{\dfrac{64x^{2}}{49y^{2}}}$

$(1\frac{7}{9})^{-1/2} =(\frac{16}{9})^{-1/2}=(\frac{4}{3})^{-1}=\frac{3}{4}$

$\sqrt { 9\displaystyle\frac { 49 }{ 64 }  } =\sqrt {\displaystyle \frac { 625 }{ 64 }  } =\displaystyle\frac { \sqrt { 625 }  }{ \sqrt { 64 }  } =\displaystyle\frac { 25 }{ 8 } =3\displaystyle\frac { 1 }{ 8 } $.

  $\sqrt{5 \left(2\dfrac{3}{4}, -, \dfrac{3}{10}\right)}$
$=\sqrt { 5\left( \dfrac { 11 }{ 4 } , -, \dfrac { 3 }{ 10 }  \right)  } $
$=\sqrt { 5\left( \dfrac { 55-6 }{ 20 } ,  \right)  } $
$ =\sqrt { \left( \dfrac { 49 }{ 4 } ,  \right)  } $
$=\dfrac{7}{2}$
$=3\dfrac{1}{2}$

$\sqrt { { 0.5 }^{ 3 }\times6\times{ 3 }^{ 5 } } \ =\sqrt { { 0.5 }^{ 2 }\times2\times0.5\times{ 3 }^{ 6 } } \ =0.5*3^{ 3 }\quad =13.5$

Given number is $ 27\dfrac {9}{16} = \dfrac {441}{16} $

Square root of $ \dfrac {441}{16} =  \dfrac { \sqrt {441}}{\sqrt {16}} = \dfrac {21}{4} = 5 \dfrac {1}{4} $

$\sqrt{\displaystyle \frac{1}{16}, +, \displaystyle \frac{1}{9}},=\sqrt{\displaystyle \frac{(1\times9)+(1\times16)}{9\times16}} =, \sqrt{\displaystyle \frac{25}{144}}, =, \displaystyle \frac{5}{12}$

$\text{Here. consider the fact that the product of 4 consecutive numbers + 1 is perfect square.}$

$So, let, x = 71$

⇒ $(71)(72)(73)(74) + 1 = x(x + 1) (x + 2) (x + 3) +1$ 

⇒$ (x² + 3x)(x² + 3x + 2) + 1$

⇒$ (x² + 3x)² + 2(x² + 3x) + 1$

⇒ $(x² + 3x + 1)^2$

⇒ $Square ,root ,of ,(x² + 3x + 1)² = x² + 3x + 1$

$Here, x = 71$

$Therefore, square root is = (71)² + (371) + 1$

⇒ $5041 + 213 + 1$ 

= $5255$

We have to find square root of $\displaystyle \frac {441}{961}$

$\displaystyle {\sqrt {\frac{0.289}{0.00121}}, =, \sqrt {\frac{0.28900}{0.00121}}, =, \sqrt {\frac{28900}{121}}}$ $=\cfrac{170}{11}$

$\sqrt{4\frac{57}{196}}$
$=\sqrt{\frac{841}{196}}$
$=\frac{29}{14}$
$=2\frac{1}{14}$

 $\sqrt{214+\sqrt{130-\sqrt{88-\sqrt{44+\sqrt{25}}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-\sqrt{44+5}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-\sqrt{49}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-7}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{81}}}$
$\Rightarrow  \sqrt{214+\sqrt{130-9}}$
$\Rightarrow \sqrt{214+\sqrt{121}}$
$\Rightarrow \sqrt{214+11}$
$\Rightarrow \sqrt{225}=15$

$\sqrt{1+2008\sqrt{1+2009\sqrt{1+2010\sqrt{1+2011.2013}}}}$
Or $\sqrt{1+2008\sqrt{1+2009\sqrt{1+2010.2012}}}$
$\Rightarrow \sqrt{1+2008\sqrt{1+2009.2011}}$
$\Rightarrow \sqrt{1+2008.2010}$
$\Rightarrow \sqrt{4036081}=2009$

Let us find the square root of $2025;and;4900$ by factorising them.
$3\mid ; ; 2025\ { \overline { 3\mid ; ; 675 }  }\ { \overline { 3\mid ; ; 225 }  }\ { \overline { 3\mid ; ; ; ; 75 }  }\ { \overline { 5\mid ; ; ; ; 25 }  }\ { \overline { 5\mid ; ; ; ; ; 5 }  }\ { \overline { ; ; \mid ; ; ; ; 1 }  }$
$2025=\underline{3\times3}\times\underline{3\times3}\times\underline{5\times5}$
$\sqrt{2025}=3\times3\times5=45$
$2\mid ; ; 4900\ { \overline { 2\mid ; ; 2450 }  }\ { \overline { 5\mid ; ; 1225 }  }\ { \overline { 5\mid ; ; ; ; 245 }  }\ { \overline { 7\mid ; ; ; ; ;49 }  }\ { \overline { 7\mid ; ; ; ; ; ;7 }  }\ { \overline { ; ; ;\mid ; ; ; ; ;1 }  }$
$4900=\underline{2\times2}\times\underline{5\times5}\times\underline{7\times7}$
$\sqrt{4900}=2\times5\times7=70$
So, $\cfrac{\sqrt{2025}}{\sqrt{4900}}=\cfrac{45}{70}$.

The 2nd , 5th and 8th digit of the number 31549786 are 196 respectively.
196 is a perfect square of 14. 
Therefore, the even number required is 14.
Second digit of that number is 4.
Answer is 4

$\sqrt{\frac{16}{49}}=\frac{n}{49}$
$\Rightarrow \frac{4}{7}=\frac{n}{49}$
$\Rightarrow n=\frac{49\times 4}{7}=28$

$15^2=225,;16^2=256$
$17^2=281;18^2=324$
$19^2=361$
$;2x-2+2x+2x+2=6(x)$
$\Rightarrow6x=324$ is possible
$\Rightarrow\sqrt{324}=18$

Given the area of the square is $441\ m^2$ .

Square root of unity is 1 and  -1 as $1^2={-1}^2=1$

To get inverse ( opposite operation) of squares. Division is opposite of Multiplication.

$13 \times 13 =169$ ( Square)

$169 \div 13 = 13$ ( Square root)

Therefore, B is the correct answer

Let $\sqrt{0.0169 \times x}=1.3$
Then,  $0.0169x = (1.3)^2 = 1.69$
$\Rightarrow x=\frac{1.69}{0.0169}=100$

$10a: \sqrt { 20a } \times \sqrt { 5a } =\sqrt{100{a}^{2}}=10a$

False

$\sqrt{10+\sqrt{25+\sqrt{121}}}$
$=\sqrt{10+\sqrt{15+11}}$
$=\sqrt{10+6}=\sqrt{16}=4$

$400=2\times 200$

$289 = 17 × 17$
$\text{So, square root of 289 is 17.}$ 

$ =\dfrac { 9\sqrt { 2\times 2\times 2\times 3 } +8\sqrt { 2\times 3\times 3\times 3 }  }{ 72 } $
$ = \dfrac { 18\sqrt { 6 } +24\sqrt { 6 }  }{ 72 } $
$ = \dfrac { 42\sqrt { 6 }  }{ 72 } $
$ = \dfrac { 7\sqrt { 6 }  }{ 12 }  $

$\cfrac { 10\sqrt { 6.25 }  }{ \sqrt { 6.25-0.5 }  } =\cfrac { 10\times 2.5 }{ 2.5-0.5 } =\cfrac { 25 }{ 2 } =12.5$

$\\225-1=224\\224-3=221\\221-5=216\\216-7=209\\209-9=200\\200-11=189\\189-13=176\\176-15=161\\161-17=144\\144-19=125\\125-21=104\\104-23=81\\81-25=56\\56-27=29\\29-29=0\\\>Total\>steps\>of\>=15\>\\hence\>\sqrt{225}=15$

   $\sqrt{\dfrac{25}{32} \times 2\dfrac{13}{18}\times 0.25}$
$=\sqrt { \dfrac { 25 }{ 32 } \times \dfrac { 49 }{ 18 } \times \dfrac { 25 }{ 100 }  } $
$=\sqrt { \dfrac { 5^ 2 }{ 2^ 5 } \times \dfrac { 7^ 2 }{ 9\times 2 } \times \dfrac { 1 }{ 2^ 2 }  } $
$=\sqrt { \dfrac { 5^ 2 }{ 2^ 6 } \times \dfrac { 7^ 2 }{ 3^2 } \times \dfrac { 1 }{ 2^ 2 }  } $
$=\dfrac{5\times 7}{2^4\times 3}$
$=\dfrac{35}{48}$

Given number is $ 96\frac {1}{25} = \frac {2401}{25} $

Square root of $ \frac {2401}{25} =  \frac { \sqrt {2401}}{\sqrt {25}}

= \frac {49}{5} = 9 \frac {4}{5} $

Calculate the square root of 1886, which is approximately 43.42. The nearest perfect square below 1886 is 43^2 = 1849. The difference is 1886 - 1849 = 37.

$\sqrt{\left (5+2\dfrac{21}{25}\right )\times \dfrac{0.169}{1.6}}\times 100$

   $\sqrt { 1\frac { 4 }{ 5 } \times 14\frac { 21 }{ 44 } \times 2\frac { 7 }{ 55 }  }$
$=\sqrt { \frac { 9 }{ 5 } \times \frac { 637 }{ 44 } \times \frac { 117 }{ 55 }  } $
$=\sqrt { \frac { 3^ 2 }{ 5 } \times \frac { 7^ 2\times 13 }{ 11\times 4 } \times \frac { 13\times 9 }{ 11\times 5 }  } $
$=\sqrt { \frac { 3^ 4\times 7^ 2\times 13^ 2 }{ 5^ 2\times 11^ 2\times 2^ 2 }  }$
$=\frac { 3^ 2\times 7\times 13 }{ 5\times 11\times 2 }$
$=\frac{819}{110}$
$=7\frac{49}{110}$

$\sqrt{100}+\sqrt{49}= 10 + 7 = 17.$

$\sqrt{42, \displaystyle \cfrac{583}{1369}}, =, \sqrt{\displaystyle \cfrac{58081}{1369}}$
$=, \displaystyle \cfrac{\sqrt{58081}}{\sqrt{1369}}$
$=, \displaystyle \cfrac{241}{37}, =, 6, \displaystyle \cfrac{19}{37}$

$\sqrt{41-\sqrt{21+\sqrt{19-\sqrt{9}}}}\=\sqrt{41-\sqrt{21+\sqrt{19-3}}}\=\sqrt{41-\sqrt{21+\sqrt{16}}}\=\sqrt{41-\sqrt{21+4}}\=\sqrt{41-\sqrt{25}}\=\sqrt{41-5}=\sqrt{36}=6$

$\sqrt {\displaystyle \frac { 25 }{ 81 } -\displaystyle\frac { 1 }{ 9 }  } =\sqrt { \displaystyle\frac { 25-9 }{ 81 }  } =\sqrt {\displaystyle \frac { 16 }{ 81 }  } =\displaystyle\frac { \sqrt { 16 }  }{ \sqrt { 81 }  } =\displaystyle\frac { 4 }{ 9 } $

$\sqrt{49}+\sqrt{0.49}+\sqrt{0.0049}+\sqrt{0.000049}\=7+\sqrt { \displaystyle\frac { 49 }{ 100 }  } +\sqrt {\displaystyle \frac { 49 }{ 10000 }  } +\sqrt { \displaystyle\frac { 49 }{ 1000000 }  } \ =7+\displaystyle\frac { \sqrt { 49 }  }{ \sqrt { 100 }  } +\displaystyle\frac { \sqrt { 49 }  }{ \sqrt { 10000 }  } +\displaystyle\frac { \sqrt { 49 }  }{ \sqrt { 1000000 }  } \ =7+\displaystyle\frac { 7 }{ 10 } +\displaystyle\frac { 7 }{ 100 } +\displaystyle\frac { 7 }{ 1000 } \ =7+0.7+0.07+0.007\ =7.777$

$\sqrt{169}$ $\rightarrow$ $13$

$\sqrt{0.000441}$

Let, $\sqrt{1+2+3+4+5+x}=l$, where $ l $ is an positive integer.
$\Rightarrow \sqrt{10+x}=l$
$\Rightarrow 10+x=l^2$
$\Rightarrow x=l^2-10$
Given, $1<x<100$
$\Rightarrow 1<l^2-10<100$
$\Rightarrow 11<l^2<110$
$\Rightarrow \sqrt{11}<l<\sqrt{110}$
$\Rightarrow 3.32<l<10.49$
$\Rightarrow l=4,5,6,7,8,9,10$              ($\because l\text{ is an integer.}$)
Therefore, there are 7 values of $x$.

For $\sqrt{100}$
$100-1=99$
$99-3=96$
$96-5=91$
$91-7=84$
$84-9=75$
$75-11=64$
$64-13=51$
$51-15=36$
$36-17=29$
$19-19=0$

$\displaystyle {\sqrt {\frac{36.1}{102.4}}, =, \sqrt {\frac{361}{1024}} = \frac{19}{32}}$

We know that the sum of the first n odd natural numbers is $n^{2}$
From $100$, we subtract successive odd numbers starting from $1$ as under
$100-1=99;;;99-3=96$
$96-5=91;;;;91-7=84$
$84-9=75;;;;75-11=64$
$64-13=51;;;51-15=36$
$36-17=19;;;19-19=0$
and obtain $0$ at $10th$ step.
$\therefore;\sqrt{100}=10$.
From $169$, we subtract successive odd numbers starting from $1$ as under
$169-1=168;;;168-3=165$
$165-5=160;;;;160-7=153$
$153-9=144;;;;144-11=133$
$133-13=120;;;120-15=105$
$105-17=88;;;88-19=69$
$69-21=48;;;48-23=25$
$25-25=0$
and obtain $0$ at $13th$ step.
$\therefore;\sqrt{169}=13$.

$\sqrt{1\displaystyle\frac{1}{2}-\begin{bmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}+\begin{pmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{4}\end{pmatrix}\end{bmatrix}}$
$\Rightarrow \sqrt{\frac{3}{2}-\left [ \frac{3}{2}-\frac{3}{2}+\left ( \frac{3}{2}-\frac{3}{2}-\frac{5}{4} \right ) \right ]}$
$\Rightarrow \sqrt{\frac{3}{2}-\left [ 0+\left ( \frac{6-6-5}{4} \right ) \right ]}$
$\Rightarrow \sqrt{\frac{3}{2}-\frac{5}{4}}$
$\Rightarrow \sqrt{\frac{6 -5}{4}}$
$\Rightarrow \sqrt{\frac{1}{4}}$
$\Rightarrow \frac{1}{2}$

$\sqrt{1, +, \displaystyle \cfrac{27}{169}}, =, \displaystyle \cfrac{196}{169}, =, \displaystyle \cfrac{14}{13}$
$\Rightarrow 1, \displaystyle \cfrac{1}{13}, =, 1, +, \displaystyle \cfrac{1}{13}$
$\therefore, x, =, 1$

$\left(\dfrac{25}{64}\right)^\frac{1}{2}$ = $\left(\dfrac{5^2}{8^2}\right)^\frac{1}{2}$ = $\dfrac{5}{8}$

$576=2^{6} \times 3^{6}=(2^{3} \times3)^2=(24)^{2}$

$144 = 2\times2\times2\times2\times 3\times3$

$ 81 = 3\times3\times3\times3$

$225 -1 = 224$ ,

$\sqrt[3]{0.000064}$$=$$0.04$

$\sqrt{\dfrac{400}{25}}$
= $\dfrac{20}{5} = 4$.

$\sqrt{\dfrac{484}{121}}$
= $\dfrac{22}{11}$
= 2

$\sqrt{841}= \sqrt{29 \times 29}$
So, the square root of 841 is 29.

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 484 - 1 = 483
483 - 3 = 480
480 - 5 = 475
475 - 7 = 468
468 - 9 = 459
459 - 11 = 448
448 - 13 = 435
435 - 15 = 420
420 - 17 = 403
403 - 19 = 384
384 - 21 = 363
363 - 23 = 340
340 - 25 = 315
315 - 27 = 288 
27 is the odd number, subtracting with 315 to get the value of 288.

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 64 - 1 = 63
63 - 3 = 60
60 - 5 = 55
55 - 7 = 48
48 - 9 = 39
39 - 11 = 28
28 - 13 = 15
15 - 15 = 0.
15 is the odd number, we get the value of zero for the square root of the number 64.

$\sqrt{\dfrac{144}{256}}$
= $\dfrac{12}{16}$.
= 0.75

$\sqrt{1369}= \sqrt{37 \times 37}$
So, the square root of 1369 is 37.

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 169 - 1 = 168
168 - 3 = 165
165 - 5 = 160
160 - 7 = 153
153 - 9 = 144
9 is the odd number, subtracting with 153 to get the value of 144.

$\sqrt{\dfrac{784}{196}}$
= $\dfrac{28}{14} = 2$.

Let $A=\sqrt { x } +\cfrac { 1 }{ \sqrt { x }  } $
$\Rightarrow { A }^{ 2 }=x+\cfrac { 1 }{ x } -2=\left( 5+2\sqrt { 6 }  \right) +\cfrac { 1 }{ 5+2\sqrt { 6 }  } -2$
$=5+2\sqrt { 6 } +\cfrac { 5-2\sqrt { 6 }  }{ 25-24 } -2=8$ $\text{[Rationalising the denominator]}$
$=5+2\sqrt { 6 } + { 5-2\sqrt { 6 }  } -2=8$
$ \Rightarrow A^2=8 $
$ \Rightarrow A=2\sqrt { 2 } $

${ ax }^{ 2 }+2b{ |x| }-c=a{ |x| }^{ 2 }+2b|x|-c$

Rupee collected from each member $=$ number of member in group $=$ $k$

If $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ is a perfect square
then $\displaystyle \frac { 1 }{ 4 } \sqrt { x } =2\times \sqrt { x } \times \left( \pm a \right) $
$\displaystyle \therefore \quad a=\pm \frac { 1 }{ 8 } $