How do you find 2 consecutive even integers whose product is 224?

2 Answers
Shwetank Mauria
Mar 27, 2016

Two consecutive even integers are {14,16}{14,16} or {-16,-14}{−16,−14}

Explanation:

Two consecutive even integers could be nn and n+2n+2 and as their product is 224224, we have

n(n+2)=224n(n+2)=224 or n^2+2n=224n2+2n=224 or

n^2+2n-224=0n2+2n−224=0

hence n=(-2+-sqrt(2^2-4*1*(-224)))/(2*1)n=−2±√22−4⋅1⋅(−224)2⋅1 or

n=(-2+-sqrt(4+896))/(2*1)=(-2+-sqrt900)/2=(-2+-30)/2n=−2±√4+8962⋅1=−2±√9002=−2±302

Hence n=(-2+30)/2=14n=−2+302=14 or n=(-2-30)/2=-16n=−2−302=−16

Hence two consecutive even integers are {14,16}{14,16} or {-16,-14}{−16,−14}

Veddesh phiϕ
Mar 27, 2016

(14,16)and(-14,-16)(14,16)and(−14,−16)

Explanation:

Let the first integer may be nn

Remember that even numbers differ in 22

So,the second number will be n+2n+2

color(purple)( :.n(n+2)=224

Use distributive property color(brown)(a(b+c)=ab+ac

rarrn^2+2n=224

rarrn^2+2n-224=0

This is a Quadratic equation (in form ax^2+bx+c=0)

Use Quadratic formula

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Where

color(red)(a=1,b=2,c=-224

rarrx=(-2+-sqrt(2^2-4(1)(-224)))/(2(1))

rarrx=(-2+-sqrt(4-(-896)))/(2)

rarrx=(-2+-sqrt(4+896))/(2)

rarrx=(-2+-sqrt(900))/(2)

rarrx=(-2+-30)/(2)

Now we have two solutions

color(indigo)((-2+30)/(2)=28/2=14

color(violet)((-2-30)/(2)=-32/2=-16

n is expressed here as x

:.n=(14and -16 )

So, the integers are color(green)((14,16)and(-14,-16)

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