the sum of the ages of two friends is 20 years. four years ago,the product of their ages in year was 48. is the situation possible? if so,determine their present age?
ruquua:
thank you
for this answer y u put greater than or lesstan symbol i am not understanding properly
We put greater or less than symbol bcoz a persons age cannot be less than 0 yes
And -48 is an imaginory no.
don't put that symbol u can only write ans plzz
OK u can also write it in that way
i am not understanding
Its ur choice
At the end u can just write that's its not possible bcoz no real roots exist
whos gona brainliest
Answers
Answered by
148
let the ages be x and y
x+y=20---------------(1)
4 years ago......
(x-4)(y-4)=48
xy-4x-4y-16=48
xy-4(x+y)=48+16
xy-4(20)=64
xy=144
x=144/y
substitute in (1)
x+y=20
144/y+y=20
144+y²=20y
y²-20y+144=0
which is impossible to be solved
so, it is impossible
x+y=20---------------(1)
4 years ago......
(x-4)(y-4)=48
xy-4x-4y-16=48
xy-4(x+y)=48+16
xy-4(20)=64
xy=144
x=144/y
substitute in (1)
x+y=20
144/y+y=20
144+y²=20y
y²-20y+144=0
which is impossible to be solved
so, it is impossible
it wrong bro because you should take 4 years ago as x-4,y-4
i want the ans i am not getting the ans so can u help plz
hello 1 thing is im not bro , 2nd is i have edited my ans.
ok
Answered by
9
Given,the sum of the ages of 2 friends =20
Let the present age of one of 2 friends be x years
Then the present age of other friend =(20-x) years
Four years ago,the age of 2 friends were(x-4) and (16-x) years respectively.
Given,product of above ages=4
(x-4) (16-x)=48
16x-x^2-64+4x=48
20x-x^2-64=48
-x^2+20x-64-48=0
-x^2+20x-112=0
x^2-20x+112=0
By comparing with ax^2+bx+c=0
we get a=1;b=-20;c=112
we know that,
b^2-4ac
(-20)^2-4(1)(112)
400-448
=-48<0(no real roots)
so we can't find the present age of 2 friends
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