Estimating square roots - class-VIII

Its unit digit is 6 and tens digit is 1

$A.$

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.

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.

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

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

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

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.

$\displaystyle \frac{1, +, \sqrt{0.01}}{1, -, \sqrt{0.1}}, =, \displaystyle \frac{1, +, 0.1}{1, -, 0.32}, =, \displaystyle \frac{1.1}{0.68}, =, 1.6$

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

$ {\cfrac{1}{\sqrt{3}} = \cfrac{1}{\sqrt{3}} \times \cfrac{\sqrt{1}}{\sqrt{3}} = \cfrac{\sqrt{1}}{3} = \cfrac{1.732}{3} = 0.577}$

$\sqrt{2, \times, \sqrt{2, \times, \sqrt{2, \times, \sqrt{2, \times, 2^{1/2}}}}}$

The sqaure root of $\sqrt 7$ is $2.646$

(i)

From square root table, Square root of 13.21 is:

 √13.21 = 3.6345

The square root of 13.21 is 3.63

(ii) From square root table, Square root of 21.97 is:

 √21.97 = 4.687

The square root of 21.97 is 4.69

$\sqrt{10}=3.16227\simeq 3.1623$
$\therefore$ The square root of $10$ correct to four places of decimal is $3.1623$

$\frac{\left ( 3\tfrac{1}{4} \right )^{4}-\left ( 4\tfrac{1}{3} \right )^{4}}{\left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2}}$

$60$ is in between two perfect squares: $49$, which is ${7}^{2}$ and $64$ which is ${8}^{2}$. The difference between $64$ and $49$ is $15$ so $60$ is little more than $\cfrac{2}{3}$ of the way toward $64$ from $49$. A reasonable estimate for $\sqrt {60}$, then would be about $7.7$ which is a little more than $\cfrac{2}{3}$ toward $8$ from $7$.

Given, $\sqrt{0.01+\sqrt{0.0064}}=x$

We know $\sqrt{\dfrac{36}{5}}=\dfrac {6}{\sqrt 5}$

$16384= 128\times 128$         (perfect square)

Let the number is $x$

$√143 = √3  × √49 = √3  × 7$
$√243 = √3  × √81 = √3  × 9$
$√143  ×√243  = 3  × 9  × 7 = 189$

$x=\sqrt{12}-\sqrt{9}$

$\begin{array}{l} =\sqrt { \sqrt [ a ]{ { { 4^{ { a^{ { a^{ 2 } } } } } }\sqrt { { 6^{ { a^{ 3 } } } }\sqrt [ { { a^{ 3 } } } ]{ { { { 12 }^{ { a^{ 6 } } } }\sqrt [ { { a^{ 4 } } } ]{ { { { 18 }^{ { a^{ 10 } } } } } }  } }  }  } }  }  \ =\sqrt { 4\times 6\times 18\times 12 }  \ =72 \ Hence, \ option, , C, , is, correct, , answer. \end{array}$

According to Question

$\quad \ \quad \quad \quad \quad \quad 22.271\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 2)496.00\quad 00\quad 00\ \quad \quad \quad -4\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 42)\quad 00\quad 96\ \quad \quad -0084\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 442)\quad 12\quad 00\ \quad \quad -\quad \quad 0884\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \quad \ 4447)\quad \quad \quad \quad \quad \quad 316\quad 00\quad \ \quad \quad \quad \quad \quad \quad \quad \quad \quad -31129\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 44541)\qquad \qquad \qquad \quad \quad \quad \quad 471\quad 00\ \qquad \qquad \qquad \qquad \qquad -\quad \quad 44541\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 2559\ Rounding\quad off\quad :\quad 22.271\ \ $

  $\displaystyle 3 \dfrac{4}{5}$
$=\dfrac{19}{5}$
$=3.8$

Therefore,
Square root of $\displaystyle 3 \frac{4}{5}=1.949$
                              $=1.95$

$\quad \ \quad \quad \quad \quad \quad 0.089\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 0)\quad 0.00\quad 80\quad 00\ \quad \quad \quad -0.00\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 8)\quad \quad 00\quad 80\ \quad \quad -\quad 0064\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 169)\quad \quad \quad 16\quad 00\ \quad \quad -\quad \quad \quad \quad 1529\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \quad \ \quad \quad \quad \quad \quad \quad \quad \quad 71\quad \quad \quad \ Rounding\quad off\quad :\quad 0.089\ \ $

$\quad \ \quad \quad \quad \quad \quad 2.287\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 2)5.\quad 20\quad 05\ \quad \quad \quad -4\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 42)01\quad 20\ \quad \quad -0084\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 448)\quad 036\quad 05\ \quad \quad -\quad \quad 3584\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad \quad \quad \quad \quad \quad 21\quad $

$\quad \ \quad \quad \quad \quad \quad 15.65\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad 15)245.00\quad 00\ \quad \quad -225\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 306)\quad 20\quad 00\ \quad \quad -1836\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 3125)164\quad 00\ \quad \quad -\quad \quad 15625\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \quad \ \quad \quad \quad \quad \quad \quad 845\quad \quad \ Rounding\quad off\quad :\quad 15.65\ \ $

$\quad \ \quad \quad \quad \quad \quad 0.254\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 2)\quad 0.06\quad 50\quad 00\ \quad \quad \quad -0.04\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad 45)\quad \quad \quad 2\quad 50\ \quad \quad -\quad \quad 225\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad 504)\quad \quad \quad 25\quad 00\ \quad \quad -\quad \quad \quad \quad 2016\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \quad \ \quad \quad \quad \quad \quad \quad \quad \quad 484\quad \quad \quad \ \ $

$\quad \ \quad \quad \quad \quad \quad 9.087\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 9)82.60\quad 00\ \quad \quad \quad -81\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 180)\quad 01\quad 60\ \quad \quad -0000\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 1808)\quad 160\quad 00\ \quad \quad -\quad \quad 12864\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \quad \ 19167)\quad \quad \quad \quad \quad 3136\quad 00\quad \ \quad \quad \quad \quad \quad \quad \quad \quad \quad -31129\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \qquad \qquad \qquad \qquad 282471\quad $

$\quad \ \quad \quad \quad \quad \quad 0.77\ \quad \quad \quad \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad 7)\quad 0.60\quad 20\ \quad \quad \quad -0.49\quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 147)\quad \quad 11\quad 20\ \quad \quad -\quad 1029\quad \quad \downarrow \ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ \quad \quad \quad \quad \quad 191\quad $

 $\displaystyle 1 \dfrac{15}{16}$

 $\displaystyle 6 \frac{7}{8}$
$=\dfrac{55}{8}$
$=6.875$

Therefore,
Square root of $\displaystyle 6 \frac{7}{8}=2.62$
                             

$\sqrt{3}=10.14889$
$\sqrt{99.35}=9.967$
$\sqrt{102.30}=10.1094$
Hence,The correct sequence of the these values matching with the above number is:1,3,2

Let 

$1^{\ast} 2 \sqrt{2}, =, \sqrt{(1)^2, +, (2 \sqrt{2}^2}, =, \sqrt{1, +, 8}, =, 3$

$\displaystyle {\sqrt {\frac{2}{9}}, =, \frac{\sqrt{2}}{\sqrt{9}}, =, \frac{\sqrt{2}}{3}, =, \frac{1.4142}{3}, =, 0.4714.}$

$\displaystyle {\sqrt{\frac{8}{3}} = \frac{\sqrt{8}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{24}}{3} = \frac{4.899}{3} = 1.633.}$ 

Let $x=\sqrt { 1+\sqrt { 1+\sqrt { 1+....... }  }  }$

$17$

$\therefore$ The square root of $17=4.123$

$1.7$

$\therefore$ The square root of $1.7=1.303$

$2.5$

$\therefore$ The square root of $2.5=1.581$

$\frac{7}{8}$

$\therefore$ The square root of $\frac{7}{8}=0.935$

$\sqrt{864}=\sqrt{2^2\times2^2\times3^2\times2\times3}=2\times2\times3\sqrt{6}=12\sqrt{6}$

For $402$

Find square root by long division method.

$\displaystyle \sqrt { 4.8\times { 10 }^{ 9 } } $

$99^2$ = 9801
$100^2$ = 10000
In between this two squares, 9805 is placed.
So the average of $\frac{99 + 100}{2}= 99.5$
Then, $99.5^2 = 9900.25$
So, $\sqrt{9805} \approx 99.05$

$22^2$ = 484
$23^2$ = 576
In between this two square numbers, 500 is placed.
So average of $\frac{22 + 23}{2}= 22.5$
Then, $22.5^2 = 506.25$
So, $\sqrt{500} \approx 22.5$

$72^2$ = 5184
$73^2$ = 5329
In between this two squares, 5245 is placed.
So average of $\frac{72 + 73}{2}= 72.5$
Then, $72.5^2 = 5256.25$
So, $\sqrt{5245} \approx 72.3$

$35^2$ = 1225
$36^2$ = 1296
In between this two squares, 1235 is placed.
So the average of $\dfrac{35 + 36}{2}= 35.5$
Then, $35.5^2 = 1260.25$

$27^2$ = 729
$28^2$ = 784
In between this two squares, 750 is placed.
So average of $\frac{27 + 28}{2}= 27.5$
Then, $27.5^2 = 756.25$
So, $\sqrt{750} \approx 27.3$

The square root of $300$ is $10\sqrt 3$

The square root of $850$ is $\sqrt {850}=\sqrt {25 \times 34}=5\sqrt {34}$

Real number $(\sqrt[3]{\sqrt{75}-\sqrt{12}})^{-2}$