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

2 Answers
Mar 27, 2016

Answer:

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

Explanation:

Two consecutive even integers could be #n# and #n+2# and as their product is #224#, we have

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

#n^2+2n-224=0#

hence #n=(-2+-sqrt(2^2-4*1*(-224)))/(2*1)# or

#n=(-2+-sqrt(4+896))/(2*1)=(-2+-sqrt900)/2=(-2+-30)/2#

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

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

Mar 27, 2016

Answer:

#(14,16)and(-14,-16)#

Explanation:

Let the first integer may be #n#

Remember that even numbers differ in #2#

So,the second number will be #n+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)#