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Find two consecutive odd integers, sum of whose squares is $290$.
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Solution: (a) Let the two consecutive odd integers be $x$ and $x + 2$
$x^2 + (x + 2)^2 = 290$
$2x^2 + 4x - 286 = 0$ or $x^2 + 2x - 143 = 0$
$(x - 11) (x + 13) = 0$
$x = 11$
Required odd integers are $11$ and $13$
$x^2 + (x + 2)^2 = 290$
$2x^2 + 4x - 286 = 0$ or $x^2 + 2x - 143 = 0$
$(x - 11) (x + 13) = 0$
$x = 11$
Required odd integers are $11$ and $13$