SIMPLIFYING SQUARE ROOTS

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The following steps will be useful to simplify any square root.

(i) Decompose the number inside the square root into prime factors.

(ii) Inside the square root, if the same number is repeated twice with multiplication, it can be taken out of the square root.

(iii) Combine the like square root terms using mathematical operations.

Example :

โˆš27 + โˆš3 - โˆš12 = โˆš(3 โ‹… 3 โ‹… 3) + โˆš3 - โˆš(2 โ‹… 2 โ‹… 3)

3โˆš3 + โˆš3 - 2โˆš3

= 2โˆš3

Solved Questions

Simplify each the following square root expressions :

Question 1 :

โˆš64 + โˆš196

Answer :

Because 64 and 196 are perfect squares, we can find the square root of 64 and 194 as shown below.

โˆš64 = โˆš(8 โ‹… 8)

โˆš64 = 8

โˆš196 = โˆš(14 โ‹… 14)

โˆš196 = 14

โˆš64 + โˆš196 = 8 + 14

= 22

Question 2 :

โˆš40 + โˆš160

Answer :

Decompose 40 and 160 into prime factors using synthetic division.

โˆš40 = โˆš(2 โ‹… 2 โ‹… 2 โ‹… 5) = 2โˆš10

โˆš160 = โˆš(2 โ‹… 2 โ‹… 2 โ‹… 2 โ‹… 2 โ‹… 5) = 4โˆš10

โˆš40 + โˆš160 :

= 2โˆš10 + 4โˆš10

= 6โˆš10

Question 3 :

2โˆš425 - 3โˆš68

Answer :

Decompose 425 and 68 into prime factors using synthetic division.

โˆš425 = โˆš(5 โ‹… 5 โ‹… 17)

โˆš425 = 5โˆš17

โˆš68 = โˆš(2 โ‹… 2 โ‹… 17)

โˆš68 = 2โˆš17

2โˆš425 - 3โˆš68 :

= 2(5โˆš17) - 3(2โˆš17)

= 10โˆš17 - 6โˆš17

= 4โˆš17

Question 4 :

โˆš243 - 5โˆš12 + โˆš27

Answer :

Decompose 243, 12 and 27 into prime factors using synthetic division.

โˆš243 = โˆš(3 โ‹… 3 โ‹… 3 โ‹… 3 โ‹… 3)  =  9โˆš3

โˆš12 = โˆš(2 โ‹… 2 โ‹… 3)  =  2โˆš3

โˆš27 = โˆš(3 โ‹… 3 โ‹… 3)  =  3โˆš3

โˆš243 - 5โˆš12 + โˆš27 :

= 9โˆš3 - 5(2โˆš3) + 3โˆš3

= 9โˆš3 - 10โˆš3 + 3โˆš3

= 2โˆš3

Question 5 :

-โˆš117 - โˆš52

Answer :

Decompose 117 and 52 into prime factors using synthetic division.

โˆš117 = โˆš(3 โ‹… 3 โ‹… 13) = 3โˆš13

โˆš52 = โˆš(2 โ‹… 2 โ‹… 13) = 2โˆš13

-โˆš117 - โˆš52 :

= -3โˆš13 - 2โˆš13

= -5โˆš13

Question 6 :

(โˆš17)(โˆš51)

Answer :

Decompose 17 and 51 into prime factors.

Because 17 is a prime number, it can't be decomposed anymore. So, โˆš17 has to be kept as it is.

โˆš51 = โˆš(3 โ‹… 17) = โˆš3 โ‹… โˆš17

(โˆš17)(โˆš51) :

(โˆš17)(โˆš3 โ‹… โˆš17)

= (โˆš17 โ‹… โˆš17)โˆš3

17โˆš3

Question 7 :

(โˆš35)(2โˆš15)

Answer :

(โˆš35)(2โˆš15)

Decompose 35 and 15 into prime factors.

โˆš35 = โˆš(5 โ‹… 7) = โˆš5 โ‹… โˆš7

โˆš15 = โˆš(5 โ‹… 3) = โˆš5 โ‹… โˆš3

(โˆš35)(2โˆš15) :

(โˆš5 โ‹… โˆš7โ‹… 2(โˆš5 โ‹… โˆš3)

= 2(โˆš5 โ‹… โˆš5)(โˆš7 โ‹… โˆš3)

= 2(5)(โˆš(7 โ‹… 3)

= 10โˆš21

Question 8 :

(14โˆš117) รท (7โˆš52)

Answer :

Decompose 117 and 52 into prime factors using synthetic division.

โˆš117 = โˆš(3 โ‹… 3 โ‹… 13)

โˆš117 = 3โˆš13

โˆš52 = โˆš(2 โ‹… 2 โ‹… 13)

โˆš52 = 2โˆš13

(14โˆš117) รท (7โˆš52) :

= 14(3โˆš13) รท 7(2โˆš13)

= 42โˆš13 รท 14โˆš13

= 42โˆš13/14โˆš13

= 3

Question 9 :

(7โˆš5)2

Answer :

(7โˆš5)= 7โˆš5 โ‹… 7โˆš5

= (7 โ‹… 7)(โˆš5 โ‹… โˆš5)

= (49)(5)

= 245

Question 10 :

(โˆš3)3 + โˆš27

Answer :

(โˆš3)3 + โˆš27 = (โˆš3 โ‹… โˆš3 โ‹… โˆš3) + โˆš(3 โ‹… 3 โ‹… 3)

= (3 โ‹… โˆš3) + 3โˆš3

= 3โˆš3 + 3โˆš3

= 6โˆš3

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