The sum of ages of two friends is 20. Four years ago , the product of their ages in year was 48 .find their present ages
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let one frnd age be X.
and another be (20-X).
. four years ago (x-4)and(20-x-4)
product of their age
(x-4)(16-x)=48
16x-x²+4x-64=48
-x²+20x -112=0
,x²-20x+112=0
by quadratic formula
(-b+-√b²-4ac)/2a
(20+-√-48)/2
(20+4√3)/2, ( 20-4√3)/2
4(5+√3)/2 , 4(5-√3)/2
x=10+2√3. ,10-2√3
present age
10+4+2√3, 10+4-2√3
14+3.46 , 14-3.46
17.46 , 10.54
Answered by
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Answer: Answer is :
let one frnd age be X.
and another be (20-X).
. four years ago (x-4)and(20-x-4)
product of their age
(x-4)(16-x)=48
16x-x²+4x-64=48
-x²+20x -112=0
,x²-20x+112=0
by quadratic formula
(-b+-√b²-4ac)/2a
(20+-√-48)/2
(20+4√3)/2, ( 20-4√3)/2
4(5+√3)/2 , 4(5-√3)/2
x=10+2√3. ,10-2√3
present age
10+4+2√3, 10+4-2√3
14+3.46 , 14-3.46
17.46 , 10.54
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