If a^2+4b^2+25c^2=2ab+10bc+5ca then prove that a=2b=5c [S Chand Class 9 ICSE Math Chapter 3 (Expansion) Exercise 3.1 Question 15 Solution ]
a2+4b2+25c2=2ab+10bc+5ca
⟹ 2a2+8b2+50c2 = 4ab+20bc+10ca [Multiplying both side by 2]
⟹ (a2-4ab+4b2 )+(4b2-20bc+25c2 )+(25c2-10ca+a2 )=0
⟹ {a2-2.a.2b+(2b)2} + {(2b)2-2.2b.5c+(5c)2} +{(5c)2 -2.5c.a+(a)2} = 0
⟹ (a-2b)2 +(2b-5c)2 +(5c-a)2=0
Sum of three square terms is zero, therefore they are individually zero.
∴ (a-2b)=0 ⟹ a= 2b —(i)
(2b-5c)=0 ⟹ 2b= 5c —(ii)
and (5c-a)=0 ⟹ 5c= a —(iii)
From (i) ,(ii) and (iii) we get,
a= 2b=5c [Proved]