Question

The sum of the squares of two
Consecutive positive even integer is 340
Find the integers.​

Answer 1

Let the two integers be x and (x + 2)

According to the question

${x}^{2} + {(x + 2)}^{2} = 340 \\ {x }^{2} + {x}^{2} + 4x + 4 = 340 \\ 2 {x}^{2} + 4x - 336 = 0 \\ {x}^{2} + 2x - 168 = 0 \\ {x}^{2} + 14x - 12x - 168 = 0 \\ x(x + 14) - 12(x + 14) = 0 \\ (x + 14)(x - 12) = 0 \\ x = - 14 \: or \: 12 \\ \\ since \: x \: is \: a \: positive \: integer \\ x = 12 \\ x + 2 = 14$

there you go :)

Answer 2

Answer:

HOW TO SOLVE?

$let \: the \: numbers \: be \: \\ {x}^{2} \: and \: {x}^{2} + 2 \\ {x}^{2} + {x}^{2} + 2 = 340 \\ 2 {x}^{2} + 2 = 340 \\ 2 {x}^{2} = 340 - 2 \\ 2 {x}^{2} = 338 \\ {x}^{2} = \frac{338}{2} \\ {x}^{2} = 169 \\ \sqrt{ {x}^{2} } = \sqrt{169} \\ x = 13 \\ the \: numbers \: are \: {13}^{2} = 169 \: and \: \\ {13}^{2} + 2 = 171$

The numbers are 169 and 171.

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