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numbers added make 10, timesed make 20
i know this is more of a maths question but there isn't a maths section and many science people are good at maths.
what two numbers when added make 10 but when timesed make 20? i learnt it a couple of years ago and i was asked it again and i couldn't rememebr it and its really anoying me now. i've tried googling it but i can't find anything.
what two numbers when added make 10 but when timesed make 20? i learnt it a couple of years ago and i was asked it again and i couldn't rememebr it and its really anoying me now. i've tried googling it but i can't find anything.
Answers
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To quote from your post
"You have two equations and two unknowns, so your problem is solvable. "
But then you didn't go on to show how to solve it!
If the original poster is interested in how to solve it then here goes:
if a + b =10 then a=10-b.
substitute in the second equation (a*b=20) for 'a' gives
(10-b)*b=20 or
10b-b^2=20
re-arranging gives:
b^2-10b+20=0
So this is a quadratic equation for b. Therefore
b=(10+/-sqrt(100-80))/2 (+/- means plus or minus)
so b=(10+/-sqrt(20))/2
b=5+/-sqrt(5) since sqrt(20)=sqrt(4)*sqrt(5)=2*sqrt(5)
So if we choose the positive square root then b=5+sqrt(5) and since a=10-b then a=5-sqrt(5)
if instead we choose the negative square root then b=5-sqrt(5) and a=5+sqrt(5).
This is clear from the fact that the 2 equations are symmetrical with respect to a and b
To quote from your post
"You have two equations and two unknowns, so your problem is solvable. "
But then you didn't go on to show how to solve it!
If the original poster is interested in how to solve it then here goes:
if a + b =10 then a=10-b.
substitute in the second equation (a*b=20) for 'a' gives
(10-b)*b=20 or
10b-b^2=20
re-arranging gives:
b^2-10b+20=0
So this is a quadratic equation for b. Therefore
b=(10+/-sqrt(100-80))/2 (+/- means plus or minus)
so b=(10+/-sqrt(20))/2
b=5+/-sqrt(5) since sqrt(20)=sqrt(4)*sqrt(5)=2*sqrt(5)
So if we choose the positive square root then b=5+sqrt(5) and since a=10-b then a=5-sqrt(5)
if instead we choose the negative square root then b=5-sqrt(5) and a=5+sqrt(5).
This is clear from the fact that the 2 equations are symmetrical with respect to a and b
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