5.5-Square Roots

5.5-Square Roots Important Formulae

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Grade 8 → Math → Squares and Square Roots → 5.5-Square Roots

5.5 - Square Roots
  • The square root of a number is a value that, when multiplied by itself, gives the original number.
  • The square root of a number $x$ is written as $\sqrt{x}$.
  • For a number $x$, if $y^2 = x$, then $y = \sqrt{x}$.
  • Perfect squares have whole number square roots (e.g., $\sqrt{16} = 4$).
  • Non-perfect squares have irrational square roots (e.g., $\sqrt{2}$ is irrational).
  • The square root of 0 is 0, i.e., $\sqrt{0} = 0$.
  • To estimate square roots of non-perfect squares, use methods like long division or approximation.

5.5 - Square Roots

The square root of a number is a value that, when multiplied by itself, gives the original number. In other words, if $x^2 = a$, then $x$ is the square root of $a$. The square root of a number is denoted as $\sqrt{a}$.

For example:

  • $\sqrt{16} = 4$ because $4 \times 4 = 16$.
  • $\sqrt{25} = 5$ because $5 \times 5 = 25$.

Key Points to Remember:

  • The square root of a perfect square is an integer. For example, $ \sqrt{36} = 6$.
  • Non-perfect square numbers do not have an integer square root, but they can have a decimal or irrational square root. For example, $ \sqrt{7} \approx 2.6457513110645906$.
  • The square root of 0 is 0, i.e., $ \sqrt{0} = 0$.
  • The square root of any positive number is always non-negative, i.e., $\sqrt{a} \geq 0$ where $a \geq 0$.

Perfect Squares:

A perfect square is a number that can be expressed as the product of an integer multiplied by itself. The square roots of perfect squares are integers. For example:

  • $\sqrt{1} = 1$
  • $\sqrt{4} = 2$
  • $\sqrt{9} = 3$
  • $\sqrt{16} = 4$
  • $\sqrt{25} = 5$
  • $\sqrt{36} = 6$
  • $\sqrt{49} = 7$
  • $\sqrt{64} = 8$
  • $\sqrt{81} = 9$

Square Roots of Fractions:

The square root of a fraction is the square root of the numerator divided by the square root of the denominator. For example:

  • $\sqrt{\frac{1}{4}} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2}$.
  • $\sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$.

Square Roots of Decimals:

To find the square root of a decimal number, convert the decimal to a fraction or use a calculator for approximation. For example:

  • $\sqrt{0.25} = 0.5$ because $0.5 \times 0.5 = 0.25$.
  • $\sqrt{1.44} = 1.2$ because $1.2 \times 1.2 = 1.44$.

Estimation of Square Roots:

If the number is not a perfect square, the square root can be estimated between two perfect squares. For example, to estimate $ \sqrt{50}$:

  • Find the perfect squares nearest to 50: $ \sqrt{49} = 7$ and $ \sqrt{64} = 8$.
  • Since $50$ is between $49$ and $64$, we estimate $ \sqrt{50} \approx 7.1$ or $7.2$.

Finding Square Roots by Prime Factorization:

To find the square root of a number using prime factorization, express the number as a product of prime factors and group the factors in pairs. For example:

  • Find the prime factorization of 36: $36 = 2^2 \times 3^2$.
  • Group the factors: $(2 \times 3)^2 = 6^2$.
  • Thus, $\sqrt{36} = 6$.

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

Solution:

What could be the possible 'one's' digits of the square root of each of the following numbers?
(i) 9801
The square root of 9801 is 99. The one's digit of the square root is 9.

(ii) 99856
The square root of 99856 is 316. The one's digit of the square root is 6.

(iii) 998001
The square root of 998001 is 999. The one's digit of the square root is 9.

(iv) 657666025
The square root of 657666025 is 25600. The one's digit of the square root is 0.

Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153
(ii) 257
(iii) 408
(iv) 441

Solution:

Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153

(ii) 257

(iii) 408

(iv) 441

Explanation:

A perfect square is a number that can be expressed as the square of an integer. For example, $16$ is a perfect square because $4^2 = 16$. If a number is not a perfect square, it cannot be written as the square of any integer.

For the numbers provided:

  • 153 is not a perfect square because a perfect square cannot have 3 at unit place.
  • 257 is not a perfect square because a perfect square cannot have 7 at unit place.
  • 408 is not a perfect square because a perfect square cannot have 8 at unit place
  • 441 is a perfect square because $21^2 = 441$.

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

Solution:

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

To find the square roots of 100 and 169 by the method of repeated subtraction, we subtract odd numbers starting from 1 until we reach 0. The number of subtractions gives us the square root of the number.

Square Root of 100:

Start with 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 = 19

19 - 19 = 0

We subtracted 10 times, so the square root of 100 is $10$.

Square Root of 169:

Start with 169:

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

We subtracted 13 times, so the square root of 169 is $13$.

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

Solution:

Find the square roots of the following numbers by the Prime Factorisation Method:


(i) 729
Prime factorisation of 729: 729 = 3 × 3 × 3 × 3 × 3 × 3 = $3^6$
Square root of 729 = $3^3 = 27$

(ii) 400
Prime factorisation of 400: 400 = 2 × 2 × 2 × 2 × 5 × 5 = $2^4 \times 5^2$
Square root of 400 = $2^2 \times 5 = 4 \times 5 = 20$

(iii) 1764
Prime factorisation of 1764: 1764 = 2 × 2 × 3 × 3 × 7 × 7 = $2^2 \times 3^2 \times 7^2$
Square root of 1764 = $2 \times 3 \times 7 = 42$

(iv) 4096
Prime factorisation of 4096: 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = $2^{12}$
Square root of 4096 = $2^6 = 64$

(v) 7744
Prime factorisation of 7744: 7744 = 2 × 2 × 2 × 2 × 7 × 7 × 7 × 7 = $2^4 \times 7^4$
Square root of 7744 = $2^2 \times 7^2 = 4 \times 49 = 196$

(vi) 9604
Prime factorisation of 9604: 9604 = 2 × 2 × 7 × 7 × 11 × 11 = $2^2 \times 7^2 \times 11^2$
Square root of 9604 = $2 \times 7 \times 11 = 2 \times 7 \times 11 = 154$

(vii) 5929
Prime factorisation of 5929: 5929 = 7 × 7 × 7 × 7 × 11 × 11 = $7^4 \times 11^2$
Square root of 5929 = $7^2 \times 11 = 49 \times 11 = 539$

(viii) 9216
Prime factorisation of 9216: 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = $2^8 \times 3^4$
Square root of 9216 = $2^4 \times 3^2 = 16 \times 9 = 144$

(ix) 529
Prime factorisation of 529: 529 = 23 × 23 = $23^2$
Square root of 529 = 23

(x) 8100
Prime factorisation of 8100: 8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = $2^2 \times 3^4 \times 5^2$
Square root of 8100 = $2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90$

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

Solution:

Find the smallest whole number to make each number a perfect square and find the square root of the obtained square number:

(i) 252:

Prime factorization of 252: $252 = 2^2 \times 3^2 \times 7$

To make it a perfect square, multiply by 7.

New number: $252 \times 7 = 1764$

Square root of 1764: $ \sqrt{1764} = 42$

(ii) 180:

Prime factorization of 180: $180 = 2^2 \times 3^2 \times 5$

To make it a perfect square, multiply by 5.

New number: $180 \times 5 = 900$

Square root of 900: $ \sqrt{900} = 30$

(iii) 1008:

Prime factorization of 1008: $1008 = 2^4 \times 3^2 \times 7$

To make it a perfect square, multiply by 7.

New number: $1008 \times 7 = 7056$

Square root of 7056: $ \sqrt{7056} = 84$

(iv) 2028:

Prime factorization of 2028: $2028 = 2^2 \times 3 \times 13^13$

To make it a perfect square, multiply by 3.

New number: $2028 \times 3 = 6084$

Square root of 6084: $ \sqrt{6084} = 78$

(v) 1458:

Prime factorization of 1458: $1458 = 2 \times 3^6$

To make it a perfect square, multiply by 2.

New number: $1458 \times 2 = 2916$

Square root of 2916: $ \sqrt{2916} = 54$

(vi) 768:

Prime factorization of 768: $768 = 2^8 \times 3$

To make it a perfect square, multiply by 3.

New number: $768 \times 3 = 2304$

Square root of 2304: $ \sqrt{2304} = 48$

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

Solution:

Question: 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

Prime factorization of 252: $252 = 2^2 \times 3^2 \times 7$

To make it a perfect square, divide by 7. The number becomes $252 \div 7 = 36$.

The square root of 36 is $6$.

(ii) 2925

Prime factorization of 2925: $2925 = 3^2 \times 5^2 \times 13$

To make it a perfect square, divide by 13. The number becomes $2925 \div 13 = 225$.

The square root of 225 is $15$.

(iii) 396

Prime factorization of 396: $396 = 2^2 \times 3^2 \times 11$

To make it a perfect square, divide by 11. The number becomes $396 \div 11 = 36$.

The square root of 36 is $6$.

(iv) 2645

Prime factorization of 2645: $2645 = 5 \times 7^2 \times 11$

To make it a perfect square, divide by 5 and 11. The number becomes $2645 \div 5 \times 11 = 49$.

The square root of 49 is $7$.

(v) 2800

Prime factorization of 2800: $2800 = 2^4 \times 5^2 \times 7$

To make it a perfect square, divide by 7. The number becomes $2800 \div 7 = 400$.

The square root of 400 is $20$.

(vi) 1620

Prime factorization of 1620: $1620 = 2^2 \times 3^4 \times 5$

To make it a perfect square, divide by 5. The number becomes $1620 \div 5 = 324$.

The square root of 324 is $18$.

The students of Class VIII of a school donated Rs. 2401 in all, for 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.

Solution:

Question: The students of Class VIII of a school donated Rs. 2401 in all, for 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.

Let the number of students in the class be $x$.

Each student donated $x$ rupees, and there are $x$ students in total. So, the total donation is:

Total donation = $x \times x = x^2$

We are given that the total donation is Rs. 2401. Therefore, we have the equation:

$x^2 = 2401$

Now, take the square root of both sides:

$x = \sqrt{2401}$

$x = 49$

Therefore, the number of students in the class is 49.

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution:

Question 8

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Let the number of rows be $x$. Since each row contains as many plants as the number of rows, the number of plants in each row will also be $x$. Therefore, the total number of plants can be represented as:

$$x \times x = 2025$$

This simplifies to:

$$x^2 = 2025$$

Taking the square root of both sides:

$$x = \sqrt{2025}$$

$$x = 45$$

So, the number of rows is $45$, and the number of plants in each row is also $45$.

Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

Problem: Find the smallest square number that is divisible by each of the numbers 4, 9, and 10.

We are asked to find the smallest square number divisible by 4, 9, and 10. To do this, we need to follow the steps below:

Step 1: Find the Least Common Multiple (LCM) of 4, 9, and 10.

First, we find the prime factorization of each number:

  • 4 = $2^2$
  • 9 = $3^2$
  • 10 = $2 \times 5$

Now, take the highest power of each prime factor:

  • Highest power of 2 = $2^2$
  • Highest power of 3 = $3^2$
  • Highest power of 5 = $5^1$

Therefore, the LCM is: $LCM = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.

Step 2: Check if the LCM is a square number.

The LCM is 180, but 180 is not a perfect square. A square number must have even powers for all prime factors. We need to adjust the powers to make it a square number.

Step 3: Make the LCM a square number.

To make 180 a square, we need to ensure that the powers of all prime factors are even:

  • For 2, the power is already even ($2^2$), so no change is needed.
  • For 3, the power is already even ($3^2$), so no change is needed.
  • For 5, the power is odd ($5^1$). To make it even, we need to multiply by $5^1$ to get $5^2$.

Thus, we multiply 180 by 5 to get a square number: $180 \times 5 = 900$.

Step 4: Conclusion

The smallest square number that is divisible by 4, 9, and 10 is 900.

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

Question: Find the smallest square number that is divisible by each of the numbers 8, 15, and 20.

To find the smallest square number divisible by 8, 15, and 20, we first need to find the Least Common Multiple (LCM) of the three numbers.

Step 1: Find the prime factorization of each number.

  • 8 = $2^3$
  • 15 = $3 \times 5$
  • 20 = $2^2 \times 5$

Step 2: Find the LCM by taking the highest powers of all primes that appear.

  • LCM = $2^3 \times 3 \times 5$ = 120

Step 3: Check if 120 is a square number.

120 is not a square number because the powers of the primes in its prime factorization are not all even.

Step 4: To make 120 a square number, we need to adjust the powers of the primes. We need to multiply by the factors required to make the powers of all primes even.

  • For $2^3$, multiply by $2^1$ to get $2^4$.
  • For $3^1$, multiply by $3^1$ to get $3^2$.
  • For $5^1$, multiply by $5^1$ to get $5^2$.

Step 5: Multiply the LCM by $2^1 \times 3^1 \times 5^1$ to make it a square number.

  • Required number = $120 \times 2 \times 3 \times 5 = 3600$

Therefore, the smallest square number that is divisible by 8, 15, and 20 is 3600.