Math, asked by amitnrw, 1 year ago

a+b+c=3
a^2+b^2+c^2=6
1/a+1/b+1/c=1

find values of a , b and c

Answers

Answered by Anonymous
101

Answer:

a ≈ 2.6474

b ≈ 0.17629 - 0.73179 i

c ≈ 0.17629 + 0.73179 i

Step-by-step explanation:

a + b + c = 3

a² + b² + c² = 6

1/a + 1/b + 1/c = 1

First and foremost :

1/a +1/b + 1/c = 1

⇒ ab + bc + ca = abc ...........( 1 )

Now :

( a + b + c )² = 3²

⇒ a² + b² + c² + 2 ( ab + bc + ca ) = 9

⇒ 6 + 2 ( ab + bc + ca ) = 9

⇒ ab + bc + ca = 3/2 = 1.5

ab + bc + ca = abc = 1.5

Now :

( 1/a + 1/b + 1/c )² = 1

⇒ 1/a² + 1/b² + 1/c² + 2 ( 1/ab + 1/bc + 1/ac ) = 1

⇒ ( a²b² + b²c² + a²c² )/a²b²c² + 2 ( a + b + c )/abc = 1

⇒ (  a²b² + b²c² + a²c² )/( 1.5 )² + 2 × 3/1.5 = 1

⇒ (  a²b² + b²c² + a²c² ) / 2.25 = 1 - 4 = - 3

⇒  a²b² + b²c² + a²c² = - 3 × 2.25

We cannot come to any conclusion if we go on like this .

So let us assume that :

x³ - ( a + b + c ) x² + ( ab + bc + ca ) x - ( abc ) = 0

Inserting all values we get :

⇒ x³ - 3 x² + 3/2 x - 3/2 = 0

⇒ 2 x³ - 6 x² + 3 x - 3 = 0

By calculating the roots in polynomial calculators we get :

x ≈ 2.6474

x ≈ 0.17629 - 0.73179 i

x ≈ 0.17629 + 0.73179 i

Replace x with a , b and c

a ≈ 2.6474

b ≈ 0.17629 - 0.73179 i

c ≈ 0.17629 + 0.73179 i


amitnrw: why imaginary numbers cannot be added to give 3 let say 1+3i , 1-2i , 1-i are three imaginary numbers and if you add them then sum = 3
Anonymous: if all were imaginary then their squares should sum negative why a^2b^2 + b^2c^2 + a^2c^2 is positive then ? on top of that 1/a + 1/b + 1/c = 1
amitnrw: i just gave you an example against the statement mentioned by you for not possible solution. Justification should be proper
Anonymous: x ≈ 2.6474
x ≈ 0.17629 - 0.73179 i
x ≈ 0.17629 + 0.73179 i ... These will be the possible answers found when u are giving the equation in polynomial roots calculator .
Anonymous: replace the values of x with a,b and c each .
amitnrw: ok , I will check tomorrow
amitnrw: Quite close a+b+c = 2.99998 1/a + 1/b + 1/c = 1.000008 , a^2 + b^2 + c^2 = 5.99985 using values mentioned by you
Anonymous: You cannot get to 3 , 1 , 6 .. it will always be like this .. that is the magic of certain cubical equations ... they cannot be calculated except by Bisection method which is not possible for a human to trial and error 100 times .. so most of these are calculated by computers .
amitnrw: Thanks for the solution
Anonymous: welcome
Answered by generalRd
64

hi

here is your answer

Plz refer to the attachment

hope it helps

BE BRAINLY@@

Attachments:

yogi2584: tq
Similar questions