Find two consecutive positive odd integers wh | vedclass

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Question

(A) Let the two consecutive positive odd integers be $x$ and $x+2$.According to the problem:$x^2 + (x+2)^2 = 290$Expanding the equation:$x^2 + x^2 + 4

Vedclass · Accepted Answer

$11, 13$

Vedclass · Answer

(A) Let the two consecutive positive odd integers be $x$ and $x+2$.According to the problem:$x^2 + (x+2)^2 = 290$Expanding the equation:$x^2 + x^2 + 4x + 4 = 290$$2x^2 + 4x - 286 = 0$Dividing by $2$:$x^2 + 2x - 143 = 0$Factoring the quadratic equation:$x^2 + 13x - 11x - 143 = 0$$x(x+13) - 11(x+13) = 0$$(x-11)(x+13) = 0$So,$x = 11$ or $x = -13$.Since the integers must be positive,we take $x = 11$.The two integers are $11$ and $11+2 = 13$.Thus,the correct option is $A$.

Find two consecutive positive odd integers whose squares have the sum $290.$

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