Square Root: If x² = y, we say that the square root of y is x and we write, √y = x.

Thus, √4 = 2, √9 = 3, √196 = 14.

Cube Root: The cube root of a given number x is the number whose cube is x. We denote the cube root of x by ³√x.

Thus, ³√8  = ³√2 x 2 x 2 = 2, ³√343 = ³√7 x 7 x 7 = 7 etc.

Note:

1.√xy = √x * √y                     2. √(x/y) = √x / √y  = (√x / √y) * (√y / √y) = √xy / y

 

SOLVED EXAMPLES:

Ex. 1. Evaluate √6084 by factorization method .

Sol.     Method: Express the given number as the product of prime factors.   Now, take the product of these prime factors choosing one out of   every pair of the same primes. This product gives the square root of the given number.  Thus, resolving 6084 into prime factors, we get:

Ex. 2. Find the square root of 1471369.                                                   

Explanation: In the given number, mark off the digits  in pairs starting from the unit’s digit. Each pair and  the remaining one digit is called a period.  Now, 1² = 1. On subtracting, we get 0 as remainder.   Now, bring down the next period. Now, trial divisor is 1 x 2 = 2 and trial dividend is 47.  So, we take 22 as divisor and put 2 as quotient.  The remainder is 3.  Next, we bring down the next period which is 13.  Now, trial divisor is 12 x 2 = 24 and trial dividend is 313. So, we take 241 as dividend and 1 as quotient. The remainder is 72. ­ Bring down the next period i.e., 69. Now, the trial divisor is 121 x 2 = 242 and the trial dividend is 7269. So, we take 3as quotient and 2423 as divisor. The remainder is then zero. Hence, √1471369 = 1213.

Ex. 3. Evaluate: √248 + √51 + √ 169 .

Sol.     Given expression = √248 + √51 + 13 = √248 + √64    = √ 248 + 8 = √256 = 16.

Ex. 4. If a * b * c = √(a + 2)(b + 3) / (c + 1), find the value of 6 * 15 * 3.

Sol.      6 * 15 * 3 = √(6 + 2)(15 + 3) / (3 + 1) = √8 * 18 / 4 = √144 / 4 = 12 / 4 = 3.

Ex. 5. Find the value of √25/16.

Sol.    √ 25 / 16   = √ 25 / √ 16 = 5 / 4

 Ex. 6. What is the square root of 0.0009?

Sol.      √0.0009= √ 9 / 1000  = 3 / 100 = 0.03.

Ex. 8. What will come in place of question mark in each of the following questions?

(i) √32.4 / ?  = 2                       (ii) √86.49 + √ 5 + ( ? )² = 12.3.

            Sol.      (i) Let √32.4 / x = 2. Then, 32.4/x = 4 <=> 4x = 32.4 <=> x = 8.1.

                          (ii) Let √86.49 + √5 + x² = 12.3.

Then, 9.3 + √5+x² = 12.3 <=> √5+x²  = 12.3 – 9.3 = 3

<=> 5 + x² = 9   <=> x² = 9 – 5= 4   <=>   x = √4 = 2.

 Ex.9. Find the value of √ 0.289 / 0.00121.

Sol.      √0.289 / 0.00121 = √0.28900/0.00121 = √28900/121 = 170 / 11.

Ex.10. If √1 + (x / 144) = 13 / 12, the find the value of x.

 Sol.      √1 + (x / 144) = 13 / 12 Þ ( 1 + (x / 144)) = (13 / 12 )² = 169 / 144

Þx / 144 = (169 / 144) – 1

Þx / 144 = 25/144 Þ x = 25.

Ex. 11. Find the value of √3 up to three places of decimal.

Sol.

Ex. 12. If √3 = 1.732, find the value of √192 – 1 √48 – √75 correct to 3 places

         of decimal.                                                                                     (S.S.C. 2004)

Sol.     √192 – (1 / 2)√48 – √75 = √64 * 3 – (1/2) √ 16 * 3  – √ 25 * 3

                                                =8√3 – (1/2) * 4√3 – 5√3

                                                =3√3 – 2√3 = √3 = 1.732

Ex. 13. Evaluate: √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

Sol.      Given exp. = √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

Now, since the sum of decimal places in the numerator and denominator under the            radical sign is the same, we remove the decimal.

\        Given exp = √(95 * 85 * 18900) / (17 * 19 * 21) = √ 5 * 5 * 900  = 5 * 30 = 150.

Ex. 14. Simplify: √ [( 12.1 )² - (8.1)²] / [(0.25)² + (0.25)(19.95)]

Sol.      Given exp. = √ [(12.1 + 8.1)(12.1 - 8.1)] / [(0.25)(0.25 + 19.95)]

=√ (20.2 * 4) /( 0.25 * 20.2)   = √ 4 / 0.25 = √400 / 25 = √16 = 4.

Ex. 15. If x = 1 + √2 and y = 1 – √2, find the value of (x² + y²).

Sol.      x² + y² = (1 + √2)² + (1 – √2)² = 2[(1)² + (√2)²] = 2 * 3 = 6.

Ex. 16. Evaluate: √0.9 up to 3 places of decimal.

Sol.                 

Ex.17. If √15 = 3.88, find the value of √ (5/3).

Sol.      √ (5/3) = √(5 * 3) / (3 * 3)  = √15 / 3 = 3.88 / 3 = 1.2933…. = 1.293.

Ex. 18. Find the least square number which is exactly divisible by 10,12,15 and 18.

Sol.      L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 2² * 3² * 5.

To make it a perfect square, it must be multiplied by 5.

Required number = (2² * 3² * 5²) = 900.

Ex. 19. Find the greatest number of five digits which is a perfect square.

(R.R.B. 1998)

Sol.

Ex. 20. Find the smallest number that must be added to 1780 to make it a perfect

square.

Sol.

Ex. 21. √2 = 1.4142, find the value of √2 / (2 + √2).

Sol.      √2 / (2 + √2) = √2 / (2 + √2) * (2 – √2) / (2 – √2) = (2√2 – 2) / (4 – 2)

= 2(√2 – 1) / 2 = √2 – 1 = 0.4142.

22. If x = (√5 + √3) / (√5 – √3) and y = (√5 – √3) / (√5 + √3), find the value of (x²  + y²).

Sol.  x = [(√5 + √3) / (√5 - √3)] * [(√5 + √3) / (√5 + √3)] = (√5 + √3)² / (5 – 3)

=(5 + 3 + 2√15) / 2 = 4 + √15.

y = [(√5 - √3) / (√5 + √3)] * [(√5 - √3) / (√5 - √3)] = (√5 – √3)² / (5 – 3)

=(5 + 3 – 2√15) / 2 = 4 – √15.

x² + y² = (4 + √15)² + (4 – √15)² = 2[(4)² + (√15)²] = 2 * 31 = 62.

Ex. 23. Find the cube root of 2744.

Sol.    Method: Resolve the given number as the product of prime factors and take the product of prime factors, choosing one out of three of the same   prime factors. Resolving 2744 as the product of    prime factors, we get:

³√2744= 2 x 7 = 14.

Ex. 24. By what least number 4320 be multiplied to obtain a number which is a perfect cube?

Sol.      Clearly, 4320 = 2³ * 3³ * 2² * 5.

To make it a perfect cube, it must be multiplied by 2 * 5² i.e,50.

GAIKWAD-PATIL GROUP OF INSTITUTIONS:

ABHA College of Engineering :

An ISO 9001:2008 certified institution. Approved by AICTE New Delhi & Govt. of Maharashtra. Affiliated to RTM Nagpur University.

We are pleased to inform you that to get admission in First Year Engineering and Second Year Engineering  at ABHA college, you can contact their Mumbai’s branch office at E-Tech Academy.  For reference, you can visit on ABHA’s official website: 

Courses – Under graduate (BE):

1. Mechanical Engineering*

2. Electrical Engineering

3. Electronics and Communication Engineering

4. Electronics Engineering

5. Computer science and Engineering

6. Information Technology

Eligibility Criteria:

For admission to First Year BE & For admission to Second Year BE.

Contact on 8898656533

For admission to M.Tech.

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