Find two consecutive even positive integers,t | vedclass

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Question

(A) Let the smaller of two consecutive even positive integers be $x$.Then,the greater even positive integer is $x+2$.According to the problem,the sum

Vedclass · Accepted Answer

$10$ and $12$

Vedclass · Answer

(A) Let the smaller of two consecutive even positive integers be $x$.Then,the greater even positive integer is $x+2$.According to the problem,the sum of their squares is $244$.$	herefore x^{2} + (x+2)^{2} = 244$$	herefore x^{2} + x^{2} + 4x + 4 = 244$$	herefore 2x^{2} + 4x + 4 - 244 = 0$$	herefore 2x^{2} + 4x - 240 = 0$Dividing the equation by $2$,we get:$	herefore x^{2} + 2x - 120 = 0$Factoring the quadratic equation:$	herefore (x+12)(x-10) = 0$$	herefore x+12 = 0$ or $x-10 = 0$$	herefore x = -12$ or $x = 10$Since the required numbers are positive integers,$x = -12$ is not possible.Therefore,$x = 10$.The smaller number is $10$ and the greater number is $10+2 = 12$.Hence,the required consecutive even positive integers are $10$ and $12$.

Find two consecutive even positive integers,the sum of whose squares is $244$.

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