Mathematics — Standard (041) — CBSE Class 10 board question

Let the two consecutive negative integers be $x$ and $x + 1$.

Given: $x^2 + (x+1)^2 = 481$

$x^2 + x^2 + 2x + 1 = 481$

$2x^2 + 2x - 480 = 0$

$x^2 + x - 240 = 0$

Factorising: $x^2 + 16x - 15x - 240 = 0$

$x(x + 16) - 15(x + 16) = 0$

$(x - 15)(x + 16) = 0$

So $x = 15$ or $x = -16$.

Since the integers are negative, $x = -16$.

The two consecutive negative integers are –16 and –15.

Verification: $(-16)^2 + (-15)^2 = 256 + 225 = 481$ ✓

Source: Chapter 4, Exercise 4.2

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• The key step is letting the integers be $x$ and $x+1$ (consecutive), then forming a quadratic from the sum-of-squares condition.
• After factorising, you get two values: $x = 15$ (positive) and $x = -16$ (negative). Since the question asks for negative integers, reject $x = 15$.
• Always verify your answer — examiners award the verification step.
• The method mirrors Exercise 4.2 Q4 (positive integers summing squares to 365), just with a negative-integer condition.