Question

CHALLENGERS
Substitute k = 2, 3 in 2k-1 (2k-1) and find for what value of k; 2k-1 (2k-1) is a perfect number
Verify 6 and 496 are perfect numbers or not​

Answer 1

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Answer 2

Question :- 1) Substitute k = 2, 3 in (2k - 1)(2k - 1) and find for what value of k, (2k - 1)(2k - 1) is a perfect number ?

2) Verify 64 96 are perfect Square numbers or not ?

Solution 1 :-

Putting k = 2 :-

→ (2k - 1)(2k - 1)

→ (2k - 1)²

→ (2*2 - 1)²

→ (4 - 1)²

→ 3²

→ 9 .

Putting k = 3 :-

→ (2k - 1)²

→ (2*3 - 1)²

→ (6 - 1)²

→ 5²

→ 25 .

Now,

→ (2k - 1)(2k - 1)

→ (2k - 1)²

using (a - b)² = a² - 2ab + b²

→ 4k² - 4k + 1

We know that, For a perfect square, the roots of the quadratic equation are real and equal.

• b² - 4ac = 0

Comparing 4k² - 4k + 1 = 0 with ax² + bx + c = 0, we get ,

• a = 4
• b = (-4)
• c = 1

Therefore,

→ b² - 4ac

→ (-4)² - 4*4*1

→ 16 - 16

→ 0 .

Hence, we can conclude that, (2k - 1)(2k - 1) is Perfect Square.

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Answer 2 :-

Prime Factors of 64 :-

→ 64 = 2 * 2 * 2 * 2 * 2 * 2 = 2² * 2² * 2²

→ 64 = (2 * 2 * 2)²

→ 64 = (8)²

So, 64 is a perfect Square of 8.

Prime Factors of 96 :-

→ 96 = 2 * 2 * 2 * 2 * 2 * 3

→ 96 = 2² * 2² * (2 * 3)

→ 96 = (2*2)² * 6

→ 96 = (4)² * 6

As , we can see Now, 6 is single or with pair.

Therefore, 96 is not a perfect Square.

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