if sum of two number is18 and their product is72 then find number
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3
two numbers are 12 and 6 respectively
Answered by
1
Given-
x+y=18
=>x=18-y....(1)
xy=72....(2)
so we put (1) in (2)
we get...(18-y)(y)=72
18y- y^2=72
-y^2+18y-72=0
y^2-18y+72=0
y^2-12y-6y+72=0
y (y-12)+6 (y-12)
(y+6)(y-12)=0
y=-6
y=12
put it in x+y= 18
-6+x=18
x=24
12+x=18
x=6
x+y=18
=>x=18-y....(1)
xy=72....(2)
so we put (1) in (2)
we get...(18-y)(y)=72
18y- y^2=72
-y^2+18y-72=0
y^2-18y+72=0
y^2-12y-6y+72=0
y (y-12)+6 (y-12)
(y+6)(y-12)=0
y=-6
y=12
put it in x+y= 18
-6+x=18
x=24
12+x=18
x=6
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