Grade 12th passAlgebra

a^2+b^2+c^2=ab+bc+ca, then find the value of (c+a)/b

prakhar

8 Years agoGrade 12th pass

2 Answers

Chandan

8 Years ago

take it as

a²+b²+c²-ab-bc-ca=0

Multiply both sides with 2, we get

2( a2 + b2 + c2 – ab – bc – ca) = 0
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
(a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) = 0
(a –b)2 + (b – c)2 + (c – a)2 = 0
Since the sum of square is zero then each term should be zero
(a –b)2 = 0, (b – c)2 = 0, (c – a)2 = 0
(a –b) = 0, (b – c) = 0, (c – a) = 0
a = b, b = c, c = a
a = b = c.

therefore c+a/b=2

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Rishi Sharma

6 Years ago

Dear Student,
Please find below the solution to your problem.

Given
a^2 + b^2 + c^2 = ab + bc + ca
a^2 + b^2 + c^2 – ab – bc – ca = 0
Multiply both sides with 2, we get
2( a^2+ b^2+ c^2– ab – bc – ca) = 0
2a^2+ 2b^2+ 2c^2– 2ab – 2bc – 2ca = 0
(a^2– 2ab + b^2) + (b^2– 2bc + c^2) + (c^2– 2ca + a^2) = 0
(a –b)^2+ (b – c)^2+ (c – a)^2= 0
Since the sum of square is zero then each term should be zero
(a –b)^2= 0, (b – c)^2= 0, (c – a)^2= 0
(a –b) = 0, (b – c) = 0, (c – a) = 0
a = b, b = c, c = a
a = b = c.
therefore (c+a)/b = (c+c)/c = 2

Thanks and Regards