18
Find two consecutive negative integers, sum of whose squares is $481$.
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Let two consecutive negative integers be $x$ and $(x + 1)$
According to given statement,
$x^2 + (x + 1)^2 = 481$ (1 Mark)
$\Rightarrow x^2 + x - 240 = 0$ (1/2 Mark)
$\Rightarrow (x + 16) (x - 15) = 0$ (1 Mark)
$x = -16$ or $15$
$x = 15$ does not satisfy the given condition.
So, required integers are $-16$ and $-15$. (1/2 Mark)
According to given statement,
$x^2 + (x + 1)^2 = 481$ (1 Mark)
$\Rightarrow x^2 + x - 240 = 0$ (1/2 Mark)
$\Rightarrow (x + 16) (x - 15) = 0$ (1 Mark)
$x = -16$ or $15$
$x = 15$ does not satisfy the given condition.
So, required integers are $-16$ and $-15$. (1/2 Mark)