The number of pairs of consecutive positive e | vedclass

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Question

(A) Let $x$ and $x+2$ be two consecutive positive even integers.Given,$x^2 + (x+2)^2 = 290$.Expanding the equation: $x^2 + x^2 + 4x + 4 = 290$.$\Right

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $x$ and $x+2$ be two consecutive positive even integers.Given,$x^2 + (x+2)^2 = 290$.Expanding the equation: $x^2 + x^2 + 4x + 4 = 290$.$\Rightarrow 2x^2 + 4x - 286 = 0$.Dividing by $2$: $x^2 + 2x - 143 = 0$.Factoring the quadratic equation: $(x + 13)(x - 11) = 0$.So,$x = -13$ or $x = 11$.Since $x$ must be a positive even integer,we discard $x = -13$.For $x = 11$,the consecutive even integer is $x+2 = 13$,which is not even.Therefore,there are no such pairs of consecutive positive even integers.The number of solutions is $0$.

The number of pairs of consecutive positive even integers such that the sum of their squares is $290$ is

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