find two consecutive whole numbers whose product is 72.
Answers
Answered by
37
x= 1st integer
x+1= 2nd integer
x(x+1) = 72
x^ + 1x + 72
x^ + 1x -72 = 0
(x+9)(x-8)
if u solve u'll get x=-9 and x=8
x can't be negative.
so x=8
x=8=1st integer
x+1=8+1=9=2nd integer.
x+1= 2nd integer
x(x+1) = 72
x^ + 1x + 72
x^ + 1x -72 = 0
(x+9)(x-8)
if u solve u'll get x=-9 and x=8
x can't be negative.
so x=8
x=8=1st integer
x+1=8+1=9=2nd integer.
amruthadeshpande:
pls mark it as brainliest....
what is meant by ^
square of that number..
thanks
Answered by
14
Let the First Number be x so second Number be x + 1
According to given Condition
x(x + 1) = 72
x² + x = 72
x² + x - 72 = 0
x² - 8x + 9x - 72 = 0
x(x - 8) +9(x - 9) = 0
(x + 9)(x - 8) = 0
x = 8 ,x cant be negative nine because it is a whole number
First No = 7
Second No. = x + 1 = 8 + 1 = 9
According to given Condition
x(x + 1) = 72
x² + x = 72
x² + x - 72 = 0
x² - 8x + 9x - 72 = 0
x(x - 8) +9(x - 9) = 0
(x + 9)(x - 8) = 0
x = 8 ,x cant be negative nine because it is a whole number
First No = 7
Second No. = x + 1 = 8 + 1 = 9
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