Question

if a+b+c=18 and a^2+b^2+C^2=122 find value of ab+bc+ca​

Answer 1

Question :-

To find the value of ---

ab + bc + ca

● When given with ---

⟹ a + b + c = 18

⟹ a^2+b^2+c^2= 122

Answer :-

● Using Identity,

(a + b + c) ^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

(a + b + c) ^2 = a^2 + b^2 + c^2 + 2( ab + bc + ca)

Putting the given values,

⟹ (18)^2 = 122 + 2( ab + bc + ca)

⟹ 324 - 122 = 2 ( ab + bc + ca)

⟹ 202 / 2 = ab + bc + ca

⟹ 101 = ab + bc + ca

■ Hence, the value of ab + bc + ca is 101 .

Hope it helps !

Answer 2

Answer:

101

Step-by-step explanation:

We know,

==>

$(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$

==>

$(18)^2=122+2(ab+bc+ca)$

==>

$324-122=2(ab+bc+ca)$

==>

2(ab+bc+ca)=202

==>

Hence ;

(ab+bc+ca)=101

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