The sum of the squares of two consecutive eve | vedclass

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(D) Let the two consecutive even numbers be $x$ and $x+2$.According to the problem,the sum of their squares is $6500$.$x^{2} + (x+2)^{2} = 6500$$x^{2}

Vedclass · Accepted Answer

$56$

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(D) Let the two consecutive even numbers be $x$ and $x+2$.According to the problem,the sum of their squares is $6500$.$x^{2} + (x+2)^{2} = 6500$$x^{2} + x^{2} + 4x + 4 = 6500$$2x^{2} + 4x - 6496 = 0$Dividing the entire equation by $2$,we get:$x^{2} + 2x - 3248 = 0$To solve this quadratic equation,we factorize it:$x^{2} + 58x - 56x - 3248 = 0$$x(x + 58) - 56(x + 58) = 0$$(x + 58)(x - 56) = 0$This gives $x = 56$ or $x = -58$.Since we are looking for positive even numbers,we take $x = 56$.The two consecutive even numbers are $56$ and $58$.The smaller number is $56$.

The sum of the squares of two consecutive even numbers is $6500.$ Which is the smaller number?

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