Question

find the smallest no.by which of the following no. must be multiplied so,that the product is a perfect square÷
i)23805
ii)12150
PLZZ ANSWER ME FAST AND THIS QUESTION IS FROM CH. SQUARE AND SQUARE ROOTS. CLASS 8.​

Answer 1

Answer:

(i) 23805

The prime factors for 23805

23805 = 3×3×23×23×5

(grouping the prime factors in equal pairs we get,)

= (3×3) × (23×23) × 5

Here 5 is left out.

So, multiply by 5 we get,

= 23805 × 5

= (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)^2

(ii) 12150

The prime factors for 12150

12150 = 2×2×2×2×3×3×5×5×2

(grouping the prime factors in equal pairs we get,)

= (2×2) × (2×2) × (3×3) × (5×5) × 2

Here, prime factor 2 is left out.

So, multiply by 2 we get,

= 12150 × 2

= (2×2) × (2×2) × (3×3) × (5×5) × (2×2)

= (2×2×3×5×2) × (2×2×3×5×2)

= 120 × 120

= (120)^2

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Answer 2

Answer:

(i) 23805 = 3 x 3 x 5 x 23 x 23

3|23805

3|7935

5|2645

2|3529

2|323

|1

Grouping 23805 into pairs of equal factors:

23805 = (3 x 3) x (23 x 23) x 5

Here, the factor 5 does not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence, the smallest number by which 23805 must be multiplied is 5.

(ii) 12150 = 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5

2|12150

3 |6075

3 |2025

3 |675

3 |225

3|75

5 |25

5 |5

|1

Grouping 12150 into pairs of equal factors:

12150 = (3 x 3 x 3 x 3) x (5 x 5) x 2 x 3

Here, 2 and 3 do not occur in pairs. To be a perfect square, every prime factor has to be in pairs.

Hence, the smallest number by which 12150 must be multiplied is 2 x 3, i.e. by 6.

HOPE IT HELPS YOU