find two consecutive even natural numbers whose product is 168
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let the two consecutive natural no. be : x and x+2
so , x * x+2 = 168
=> x^2 + 2x = 168
=> x^2 +2x - 168 = 0
=> x^2 - 14x -12x - 168 (by splitting the middle term)
=> x(x+14) -12 (x+14)
from the above we get , => (x-12) = 0 ; (x+14) = 0
therefore => x=12 , x= -14
since, negative term is negligible
so x=12 is accepted
IF, x = 12 and x+ 2 = 12+2= 14
so, two consecutive even natural numbers whose product is 168 =14,12
NOTE : x^2 is x square
and (*) is multiplication symbol
HOPE YOU WILL LIKE IT.............. BEST OF LUCK
so , x * x+2 = 168
=> x^2 + 2x = 168
=> x^2 +2x - 168 = 0
=> x^2 - 14x -12x - 168 (by splitting the middle term)
=> x(x+14) -12 (x+14)
from the above we get , => (x-12) = 0 ; (x+14) = 0
therefore => x=12 , x= -14
since, negative term is negligible
so x=12 is accepted
IF, x = 12 and x+ 2 = 12+2= 14
so, two consecutive even natural numbers whose product is 168 =14,12
NOTE : x^2 is x square
and (*) is multiplication symbol
HOPE YOU WILL LIKE IT.............. BEST OF LUCK
manusrimanjari:
thank u ^-^
Answered by
2
let natural no. be x and x+2
x(x+2)=168
x^2+2x-168=0
x^2+14x-12x-168=0
x(x+14)-12(x+14)
(x-12)(x+14)
x=12
x= -14
minus is negligible
so, x=12
x+2=12+2=14
x(x+2)=168
x^2+2x-168=0
x^2+14x-12x-168=0
x(x+14)-12(x+14)
(x-12)(x+14)
x=12
x= -14
minus is negligible
so, x=12
x+2=12+2=14
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