Question

A+b+c=9, and ab+bc+ca=14, find a2+b2+c2

Answer 1

☞☞☞ Answer ☜☜☜

♦ Given :

a + b + c = 9

ab + bc + ca = 14

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Required to find ✍:

$1. \: {a}^{2} + { b}^{2} + {c}^{2}$

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✮ Identity Used :

$(x + y + z {)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx$

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✤ solution :

In the question the given information is ;

a + b + c = 9

ab + bc + ca = 14

Now, Let's consider

a + b + c = 9

➟Now let's do squaring on both sides

( a + b + c )^2 = (9)^2

a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 81

Now let's take 2 as common on the L.H.S. side

a^2 + b^2 + c^2 + 2 ( ab + bc + ca ) = 81

Hence ;

From the above the above given information

we can substitute that ;

ab + bc + ca = 14

Therefore; by substitution we get

a^2 + b^2 + c^2 + 2 ( 14 ) = 81

a^2 + b^2 + c^2 + 28 = 81

==> a^2 + b^2 + c^2 = 81 - 28

==> a^2 + b^2 + c^2 = 53

Hence ;

$\bold{\large{\boxed{\longrightarrow{\boxed{{a}^{2}+{b}^{2}+{c}^{2}\:=\:53}}}}}$

$\rule{400}{6}$