Question

Find the two consecutive positive even integers whose product is 48.

Answer 1

`6,8`

Explanation:

Method 1:

`\mathrm{Let\ }2x\ \mathrm{and\ }2x+2\ \mathrm{be\ the\ two\ consecutive\ even\ integers.}\\\mathrm{Then,}\\2x(2x+2)=48\\\mathrm{or,\ }4x^2+4x=48\\\mathrm{or,\ }x^2+x=12\\\mathrm{or,\ }x^2+x-12=0\\\mathrm{or,\ }x^2+4x-3x-12=0\\\mathrm{or,\ }x(x+4)-3(x+4)=0\\\mathrm{or,\ }(x-3)(x+4)=0\\\mathrm{i.e.}\ x=3\ \mathrm{or\ }-4.\\x=-4\mathrm{\ is\ not\ true\ because\ }2x\ \mathrm{and}\ 2x+2\ \mathrm{must\ be\ positive.}\\\mathrm{Therefore\ the\ numbers\ are:}\ 2x= 2(3)=6\ \mathrm{and\ }2x+2=6+2=8.`

Method 2:

`\mathrm{Let}\ x\ \mathrm{and}\ (48)/(x)\ \mathrm{be\ the\ required\ even\ integers.\ They\ are\ consecutive,\ therefore:}\\x=(48)/(x)+2\\\mathrm{or,\ }x=(48+2x)/(x)\\\mathrm{or,\ }x^2=48+2x\\\mathrm{or,\ }x^2-2x-48=0\\\mathrm{or,\ }x^2+6x-8x-48=0\\\mathrm{or,\ }x(x+6)-8(x+6)=0\\\mathrm{i.e.\ }x=8\ \mathrm{or\ }-6\\`

`x\ \mathrm{cannot\ be\ -6\ because\ }x\ \mathrm{is\ positive\ according\ to\ our\ supposition.}\\\mathrm{\therefore First\ number,\ }x=8\ \mathrm{and\ second\ number=}(48)/(x)=(48)/(8)=6\\`

Method 3:

`\mathrm{Let\ }x\ \mathrm{and}\ y\ \mathrm{be\ the\ two\ required\ numbers.}\\\mathrm{As\ they\ are\ consecutive\ even\ numbers,}\\x=y+2....(1)\\\mathrm{The\ product\ of\ these\ numbers\ is\ 48,\ therefore,}\\xy=48.....(2)\\\mathrm{Substituting\ }x=y+2\ \mathrm{in\ equation(2),}\\y(y+2)=48\\\mathrm{or,\ }y^2+2y-48=0\\\mathrm{or,\ }y^2+8y-6y-48=0\\\mathrm{or,\ }y(y+8)-6(y+8)=0\\\mathrm{or,\ }(y-6)(y+8)=0\\\mathrm{i.e.\ }y=6\ \mathrm{or\ }-8.\\`

`\mathrm{As}\ x\ \mathrm{and\ }\ y\ \mathrm{are\ positive\ integers,\ }y\ \mathrm{cannot\ be\ -8.\ Therefore,\ }y=6.\\\mathrm{When\ }y=6,\ x=y+2=6+2=8.\\\mathrm{So\ the\ two\ numbers\ are\ 6\ and\ 8.}`

You may consider any one number greater than the other in method 2 and 3.

Answer 2

Answer: 6 and 8

Explanation: