Question

the product of two consecutive even integers is 224 .find the integers
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Answer 1

Heyamate!!

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$\mathfrak{\green{\huge{Answer}}}$

$let \: the \: two \: number \: be \: x \: and \: x + 2 \\ so \: acco \: to \: question \\ {x}^{2} + (x + 2) {}^{2} = 224 \\ {x}^{2} + {x}^{2} + 4 + 4x = 224 \\ 2x {}^{2} + 4x + 4 - 224 = 0 \\ 2x {}^{2} + 4x - 220 = 0 \\ x {}^{2} + 2x - 110 = 0$
So x=14

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$\mathcal\red\huge {Hope\:it\:helps}$

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$\mathbb{\huge{R-K}}$

Answer 2

Answer:

let 1st integer be x then the 2nd will be

x+2

so, x(x+2) = 224

x^2+2x= 224

x^2 + 2x-224=0

x^2+16x-14x-224=0

x(x+16) - 14(x+16)=0

(x-14) (x+16)

So, x= 14 or -16

So, integers are 14 and 16 or -16 and -14