Question

find two consecutive odd positive integers some of whose squares is 290

Answer 1

Let the two consecutive odd natural no.s be x , x + 2

By the given condition :

${(x)}^{2} + {(x + 2)}^{2} = 290$


${x}^{2} + {x}^{2} + 4x + 4 = 290$

$2x ^{2} + 4x + 4 = 290$

$2 {x}^{2} + 4x + 4 - 290 = 0$

$2 {x}^{2} + 4x - 286 = 0$
[Dividing throughout by 2]

${x}^{2} + 2x - 143 = 0$

${x}^{2} + 13x - 11x - 143 = 0$

$x(x + 13) - 11(x + 13) = 0$

$(x + 13)(x - 11) = 0$
$x + 13 = 0 \: \: or \: \: x - 11 = 0$

$x = - 13 \: \: or \: \: x = 11$

But , x = -13 is not acceptable .
So,
$x = 11$
Thus,
The two consecutive natural no. are:

x = 11
x + 2 = 11 + 2 = 13
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Final answer:
The two consecutive odd positive integers are 11 & 13.
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$hope \: this \: helps$

Answer 2

Here's your solution!