Find two consecutive odd positive integers,th | vedclass

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Question

(A) Let the smaller of the two consecutive odd positive integers be $x$. Then,the second integer will be $x+2$.According to the question,$x^{2} + (x+2

Vedclass · Accepted Answer

$11, 13$

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(A) Let the smaller of the two consecutive odd positive integers be $x$. Then,the second integer will be $x+2$.According to the question,$x^{2} + (x+2)^{2} = 290$$x^{2} + x^{2} + 4x + 4 = 290$$2x^{2} + 4x - 286 = 0$Dividing by $2$,we get:$x^{2} + 2x - 143 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:$x = \frac{-2 \pm \sqrt{2^{2} - 4(1)(-143)}}{2(1)}$$x = \frac{-2 \pm \sqrt{4 + 572}}{2} = \frac{-2 \pm \sqrt{576}}{2} = \frac{-2 \pm 24}{2}$$x = \frac{22}{2} = 11$ or $x = \frac{-26}{2} = -13$Since $x$ must be a positive integer,we take $x = 11$.The two consecutive odd integers are $11$ and $13$.Verification: $11^{2} + 13^{2} = 121 + 169 = 290$.

Find two consecutive odd positive integers,the sum of whose squares is $290$.

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