Q) Find two consecutive negative integers, sum of whose squares is 481.

(Q 31 – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Let’s consider the integers are n and (n + 1)

(note: since these are consecutive, the difference will be 1)

Step 1: Given condition is “Sum of whose squares is 481.

∴ n ² + (n + 1) ² = 481

∴ n ² + n ² + 1 + 2 n = 481

∴ 2 n ² + 2 n = 481 – 1 = 480

∴ n ² + n = 240

∴ n ² + n – 240 = 0

By mid-term splitting, we get:

∴ n ² + 16 n – 15 n – 240 = 0

∴ n (n + 16) – 15 (n + 16) = 0

∴ (n + 16) (n – 15) = 0

Therefore, we get n = – 16 and n = 15

Step 2: Since we need to take negative value of n, hence we reject n = 15

And accept, n = – 16

∴ Second integer = n + 1 = – 16 + 1 = – 15

Therefore, the two consecutive negative integers are – 15 and – 16.

Check: If two integers are – 15 and – 16,
Sum of the squares = (-15)² + (-16)² = 225 + 256 = 481
Since given condition is matched, our answer is correct.

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