# NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots  Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots Exercise 6.1

1.What will be the unit digit of the squares of the following numbers?
(i) 81

(ii) 272

(iii) 799

(iv) 3853

(v) 1234

(vi) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

As we know that unit's place is affected only by the unit’s place of the number of which square is seen. So we have: (i) Unit digit of ${81}^{2}$ = 1 [Unit’s place=1 i.e, $1×1$ =1]
(ii) Unit digit of ${272}^{2}$ = 4 [Unit’s place=2 i.e, $2×2$ =4]

(iii) Unit digit of ${799}^{2}$ = 1 [Unit’s place=1 i.e, $9×9$ =1]

(iv) Unit digit of ${3853}^{2}$ = 9 [Unit’s place=3 i.e, $3×3$ =9]

(v) Unit digit of ${1234}^{2}$ = 6 [Unit’s place=4 i.e, $4×4$ =6]

(vi) Unit digit of ${26387}^{2}$ = 9 [Unit’s place=7 i.e, $7×7$ =9]

(vii) Unit digit of ${52698}^{2}$= 4 [Unit’s place=8 i.e, $8×8$ =4]

(viii) Unit digit of ${99880}^{2}$ = 0 [Unit’s place=0 i.e, $0×0$ =0]

(ix) Unit digit of ${12796}^{2}$ = 6 [Unit’s place=6 i.e, $6×6$ =6]

(x) Unit digit of ${55555}^{2}$ = 5 [Unit’s place=5 i.e, $5×5$ =5]

2. The following numbers are not perfect squares. Give reason.
(i) 1057

(ii) 23453

(iii) 7928

(iv) 222222

(v) 64000

(vi) 89722

(vii) 222000

(viii) 505050

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

For checking a number to be perfect square we need to see the unit's digit or the number of digits repeating from the end of the number given.If the unit's digit is perfect square then it may be a perfect square.Also if the number of digits repeating are odd in number then it won't be a perfect square.So here we have:
(i) It is not a perfect square number because 1057 ends with 7 at unit place.

(ii) It is not a perfect square number because 23453 ends with 3 at unit place.

(iii) It is not a perfect square number because 7928 ends with 8 at unit place.

(iv) It is not a perfect square number because 222222 ends with 2 at unit place.

(v) It is not a perfect square number because 64000 ends with odd number of zeros i.e, 3.

(vi) It is not a perfect square number because 89722 ends with 2 at unit place.

(vii) It is not a perfect square number because 22000 ends with 3 zeros.

(viii) It is not a perfect square number because 505050 ends with 1 zero.

3.The squares of which of the following would be odd numbers?
(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

The squares of a number would be odd number depending upon the unit's digit of the number.If the number has an even unit digit then it will have an even square number and if the unit's digit is odd then it will have an odd square number.Here we have:
(i) ${431}^{2}$ is an odd number as the unit's digit is odd.

(ii) ${2826}^{2}$ is an even number as the unit's digit is even.

(iii) ${7779}^{2}$ is an odd number as the unit's digit is odd.

(iv) ${82004}^{2}$ is an even number as the unit's digit is even.

4.Observe the following pattern and find the missing digits.
${11}^{2}$ = 121

${101}^{2}$ = 10201

${1001}^{2}$ = 1002001

${100001}^{2}$ = 1…2…1

${10000001}^{2}$ = ………

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

According to the above pattern, we can observe that the squares of the given numbers have the same number of zeroes before and after the digit 2.Therefore, we have
${100001}^{2}$ = 10000200001

${10000001}^{2}$ = 100000020000001

5.Observe the following pattern and supply the missing numbers.
${11}^{2}$ = 121

${101}^{2}$ = 10201

${10101}^{2}$ = 102030201

${1010101}^{2}$ = ……….

$\dots \dots \dots {.}^{2}$ = 10203040504030201

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

According to the above pattern, we have
${1010101}^{2}$ = 1020304030201

${101010101}^{2}$ = 10203040504030201

6.Using the given pattern, find the missing numbers.
${1}^{2}+{2}^{2}+{2}^{2}={3}^{2}$

${2}^{2}+{3}^{2}+{6}^{2}={7}^{2}$

${3}^{2}+{4}^{2}+{12}^{2}={13}^{2}$

${4}^{2}+{5}^{2}+\dots {.}^{2}={21}^{2}$

${5}^{2}+\dots {.}^{2}+{30}^{2}={31}^{2}$

${6}^{2}+{7}^{2}+\dots .{.}^{2}=\dots \dots 2$

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

According to the given pattern, we have the third number as the product of the first two numbers and fourth number is obtained after adding 1 to the third number.

${4}^{2}+{5}^{2}+{20}^{2}={21}^{2}$

${5}^{2}+{6}^{2}+{30}^{2}={31}^{2}$

${6}^{2}+{7}^{2}+{42}^{2}={43}^{2}$

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

We know that the sum of n odd numbers =${n}^{2}$.So, we have:

(i) 1 + 3 + 5 + 7 + 9 = $\left(5{\right)}^{2}$ = 25 [? n = 5]

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100 [? n = 10]

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = $\left(12{\right)}^{2}$ = 144 [? n = 12]

8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

(i) 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)

(ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

9.How many numbers lie between squares of the following numbers?
(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100.

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

We know that numbers between ${n}^{2}$ and $\left(n+1{\right)}^{2}$ = 2n

(i) Numbers between ${12}^{2}$ and ${13}^{2}$ = (2n) = $2×12$ = 24

(ii) Numbers between ${25}^{2}$ and ${26}^{2}$ = $2×25$ = 50 [ n = 25]

(iii) Numbers between ${99}^{2}$ and ${100}^{2}$ = $2×99$ = 198 [ n = 99]

## NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots Exercise 6.2

1.Find the square of the following numbers.
(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

(i) We have, $32=30+2$

$\left(32{\right)}^{2}=\left(30+2{\right)}^{2}$

$=30\left(30+2\right)+2\left(30+2\right)$

$={30}^{2}+30×2+2×30+{2}^{2}$

$=900+60+60+4$

$=1024$

So we have $\left(32{\right)}^{2}=1024$

(ii)We have, $35=\left(30+5\right)$

$\left(35{\right)}^{2}=\left(30+5{\right)}^{2}$

$=30\left(30+5\right)+5\left(30+5\right)$

$=\left(30{\right)}^{2}+30×5+5×30+\left(5{\right)}^{2}$

$=900+150+150+25$

$=1225$

So we have $\left(35{\right)}^{2}=1225$

(iii) We have, $86=\left(80+6\right)$

${86}^{2}=\left(80+6{\right)}^{2}$

$=80\left(80+6\right)+6\left(80+6\right)$

$=\left(80{\right)}^{2}+80×6+6×80+\left(6{\right)}^{2}$

$=6400+480+480+36$

$=7396$

So we have, $\left(86{\right)}^{2}=7396$

(iv)We have, $93=\left(90+3\right)$

${93}^{2}=\left(90+3{\right)}^{2}$

$=90\left(90+3\right)+3\left(90+3\right)$

$=\left(90{\right)}^{2}+90×3+3×90+\left(3{\right)}^{2}$

$=8100+270+270+9$

$=8649$

So we have, $\left(93{\right)}^{2}=8649$

(v) We have, $71=\left(70+1\right)$

${71}^{2}=\left(70+1{\right)}^{2}$

$=70\left(70+1\right)+1\left(70+1\right)$

$=\left(70\right)2+70×1+1×70+\left(1\right)2$

$=4900+70+70+1$

$=5041$

So we have, $\left(71{\right)}^{2}=5041$

(vi)We have, $46=\left(40+6\right)$

${46}^{2}=\left(40+6{\right)}^{2}$

$=40\left(40+6\right)+6\left(40+6\right)$

$=\left(40{\right)}^{2}+40×6+6×40+\left(6{\right)}^{2}$

$=1600+240+240+36$

$=2116$

So we have, $\left(46{\right)}^{2}=2116$

2.Write a Pythagorean triplet whose one member is
(i) 6

(ii) 14

(iii) 16

(iv) 18

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

For a natural number m greater than 1,Triplets are in the form 2m, m2 – 1, m2 + 1
(i) Let ${m}^{2}–1=6$

$?{m}^{2}=6+1=7$

So,here we observe that value of m is not integer.

Now, let us try for ${m}^{2}+1=6$

$?{m}^{2}=6–1=5$

So, here too we have no integer value of m.

Now, let 2m = 6

$?m=3\phantom{\rule{0.278em}{0ex}}which\phantom{\rule{0.278em}{0ex}}is\phantom{\rule{0.278em}{0ex}}an\phantom{\rule{0.278em}{0ex}}integer.$

Other number are:

${m}^{2}–1=32–1=8$ and ${m}^{2}+1=32+1=10$

Hence, the required triplets are 6, 8 and 10.

(ii) Let ${m}^{2}–1=14$

$?{m}^{2}=1+14=15$

So, we observe that the value of m will not be an integer.

Now take 2m = 14

$?$ m = 7 which is an integer.

The other numbers of triplets are 2m =$2×7$ = 14

${m}^{2}–1=\left(7\right)2–1=49–1=48$

and ${m}^{2}+1=\left(7\right)2+1=49+1=50$

Hence the triplets are 14, 48 and 50.

(iii) Let, 2m = 16

$?$m = 8

The other numbers are

${m}^{2}–1=\left(8\right)2–1=64–1=63$

${m}^{2}+1=\left(8\right)2+1=64+1=65$

Hence the triplets are 16, 63 and 65

(iv) Let 2m = 18

$?$ m = 9

The other triplets are:

${m}^{2}–1=\left(9\right)2–1=81–1=80$

and ${m}^{2}+1=\left(9\right)2+1=81+1=82$

Hence we have the triplets as 18, 80 and 82.

## NCERT solutions for class 8 Maths Chapter 6 Squares And Square Roots Exercise 6.3

1.What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801

(ii) 99856

(iii) 998001

(iv) 657666025

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

One’s digit of any number depends on the one’s digit of the number of which it is square of. Here we have:
(i) One’s digit in the square root of 9801 can be 1 or 9.

(ii) One’s digit in the square root of 99856 can be 4 or 6.

(iii) One’s digit in the square root of 998001 can be 1 or 9.

(iv) One’s digit in the square root of 657666025 can be 5.

2.Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153

(ii) 257

(iii) 408

(iv) 441

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares because they don't exist at the end of any square number.
(i) 153 is not a perfect square number because it ends with 3.

(ii) 257 is not a perfect square number because it ends with 7.

(iii) 408 is not a perfect square number because it ends with 8.

(iv) 441 is a perfect square number.

3.Find the square roots of 100 and 169 by the method of repeated subtraction.

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

Using the method of repeated subtraction of consecutive odd numbers, we have

(i) 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

Thus it takes 10 times repetition to reach 0.

So, $\sqrt{100}=10$

(ii) 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

It takes thirteen times repetition to reach 0.

Hence, $\sqrt{169}=13$

4.Find the square roots of the following numbers by the prime factorisation Method. (i) 729 (ii) 400 (iii) 1764 (iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(viii) 9216

(ix) 529

(x) 8100

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

(i) We have prime factors of 729 as:

$\begin{array}{cl}3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$729=3×3×3×3×3×3={3}^{2}×{3}^{2}×{3}^{2}$

?$\sqrt{729}=3×3×3=27$

(ii) We have prime factors of 400 as

$\begin{array}{cl}2& 400\\ 2& 200\\ 2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$400=2×2×2×2×5×5={2}^{2}×{2}^{2}×{5}^{2}$

?$\sqrt{400}=2×2×5=20$

(iii) We have prime factors of 1764 as:

$\begin{array}{cl}2& 1764\\ 2& 882\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$1764=2×2×3×3×7×7={2}^{2}×{3}^{2}×{7}^{2}$

?$\sqrt{1764}=2×3×7=42$

(iv) We have prime factors of 4096 as:

$\begin{array}{cl}2& 4096\\ 2& 2048\\ 2& 1024\\ 2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$4096=2×2×2×2×2×2×2×2×2×2×2×2$

$={2}^{2}×{2}^{2}×{2}^{2}×{2}^{2}×{2}^{2}×{2}^{2}$

?$\sqrt{4096}=2×2×2×2×2×2=64$

(v)We have prime factors of 7744 as:

$\begin{array}{cl}2& 7744\\ 2& 3872\\ 2& 1936\\ 2& 968\\ 2& 484\\ 2& 242\\ 11& 121\\ 11& 11\\ & 1\end{array}$

$7744=2×2×2×2×2×2×11×11$

$={2}^{2}×{2}^{2}×{2}^{2}×{11}^{2}$

?$\sqrt{7744}=2×2×2×11=88$

(vi) We have prime factors of 9604 as:

$\begin{array}{cl}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$9604=2×2×7×7×7×7={2}^{2}×{7}^{2}×{7}^{2}$

?$\sqrt{9604}=2×7×7=98$

(vii) We have prime factors of 5929 as:

$\begin{array}{cl}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}$

$5929=7×7×11×11={7}^{2}×{11}^{2}$

?$\sqrt{5929}=7×11=77$

(viii) We have prime factors of 9216 as:

$\begin{array}{cl}2& 9216\\ 2& 4608\\ 2& 2304\\ 2& 1152\\ 2& 576\\ 2& 288\\ 2& 144\\ 2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$9216=2×2×2×2×2×2×2×2×2×2×3×3$

$={2}^{2}×{2}^{2}×{2}^{2}×{2}^{2}×{2}^{2}×{3}^{2}$

?$\sqrt{9216}=2×2×2×2×2×3=96$

(ix) We have prime factors of 529 as:

$\begin{array}{cl}23& 529\\ 23& 23\\ & 1\end{array}$

$529=23×23={23}^{2}$

?$\sqrt{529}=23$

(x)We have prime factors of 8100 as:

$\begin{array}{cl}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$8100=2×2×3×3×3×3×5×5={2}^{2}×{3}^{2}×{3}^{2}×{5}^{2}$

?$\sqrt{8100}=2×3×3×5=90$

5.For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.
(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

(i) We have prime factors of 252 as:
$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$

$252=2×2×3×3×7$

Here, the prime factors are not in pair as 7 has no pair. Thus, the smallest whole number by which the given number is to be multiplied to get a perfect square number is 7.

The new square number is $252×7=1764$

?Square root of 1764 is

$\sqrt{1764}=2×3×7=42$

(ii)We have prime factors of180 as:

$\begin{array}{cl}2& 180\\ 2& 90\\ 5& 45\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$180=2×2×3×3×5$

Here, 5 has no pair.So the least number to be multiplied to get perfect square is 5

New square number = $180×5=900$

The square root of 900 is

?$\sqrt{900}=2×3×5=30$

(iii) We have prime factors of 1008 as:

$\begin{array}{cl}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$

$1008=2×2×2×2×3×3×7$

Here, 7 has no pair.So the least number to be multiplied to get perfect square is 7

New square number = $1008×7=7056$

Square root of 7056 is

?$\sqrt{7056}=2×2×3×7=84$

(iv) We have prime factors of 2028 as:

$\begin{array}{cl}2& 2028\\ 2& 1014\\ 3& 507\\ 13& 169\\ 13& 13\\ & 1\end{array}$

$2028=2×2×3×13×13$

Here, 3 is not in pair.So the least number to be multiplied to get perfect square is 3

New square number =$2028×3=6084$

Square root of 6084 is

?$\sqrt{6084}=2×13×3=78$

(v)We have prime factors of 1458 as:

$\begin{array}{cl}2& 1458\\ 2& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$1458=2×3×3×3×3×3×3$

Here, 2 is not in pair.So the least number to be multiplied to get perfect square is 2

New square number =$1458×2=2916$

Square root of 2916 is

?$\sqrt{2916}=3×3×3×2=54$

(vi) We have prime factors of 768 as:

$\begin{array}{cl}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$

$768=2×2×2×2×2×2×2×2×3$

Here, 3 is not in pair.So the least number to be multiplied to get perfect square is 3

New square number = $768×3=2304$

Square root of 2304 is

?$\sqrt{2304}=2×2×2×2×3=48$

6.For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.
(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

(i) We have prime factors of 252 as:
$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$

$252=2×2×3×3×7$

Here, the prime factors are not in pair as 7 has no pair. Thus, the smallest whole number by which the given number is to be divided to get a perfect square number is 7.

The new square number is $252÷7=36$

?Square root of 36 is $\sqrt{36}=6$

(ii) We have prime factors of 2925 as:
$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}$

$2925=3×3×5×5×13$

Here, 13 has no pair.So, the smallest whole number by which 2925 is divided to get a square number is 13.

New square number =$2925÷13=225$

? $\sqrt{225}=15$

(iii)We have prime factors of 396 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}$

$396=2×2×3×3×11$

Here 11 is not in pair.So,the required smallest whole number by which 396 is divided to get a square number is 11.

New square number = $396÷11=36$

? $\sqrt{36}=6$

(iv)We have prime factors of 2645 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}$

$2645=5×23×23$

Here, 5 is not in pair.So, 5 is the required smallest whole number by which 2645 is divided to get a square number

New square number = $2645÷5=529$

? $\sqrt{529}=23$

(v) We have prime factors of 2800 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}$

$2800=2×2×2×2×5×5×7$

Here, 7 is not in pair.So, 7 is the required smallest number by which 2800 is divided to get a square number.

New square number =$2800÷7=400$

? $\sqrt{400}=20$

(vi)We have prime factors of 1620 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$

$1620=2×2×3×3×3×3×5$

Here, 5 is not in pair.So, 5 is the required smallest prime number by which 1620 is divided to get a square number = $1620÷5=324$

? $\sqrt{324}=18$

7.The students of class VIII of a school donated ? 2401 in all, for the Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

NCERT Solutions for Class 8 Maths Chapter 6 Squares And Square Roots

Let x be the number of students in the class. Therefore, total money donated by students:$x×x=?{x}^{2}$
Here we have, ${x}^{2}=2401$