TS ICET 23 May 2018 Shift 1 Paper

Question : 123 of 200

Marks: +1, -0

a + b + c = 9;

a^2 + b^2 + c^2

= 27 ⇒ab + be + ca

Solution:

(a+b+c)^{2}=(a^{2}+b^{2}+c^{2})+2(a b+b c+c a)

⇒(a b+b c+c a)=[(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})]{/}2

=[9^{2}-27]{/}2=27

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