Question

LE a+b+c=9 and ab+bc+ca = 23 Find
the value
of a2+b² +c²
2
0​

Answer 1

GiveN :-

• $\sf \: a + b + c = 9 \\ \\ \sf \: ab + bc + ca = 23$

To FinD :-

• $\sf \: the \: value \: of \: ({a}^{2} + {b}^{2} + {c}^{2} )$

SolutioN :-

$\implies\sf \: {a}^{2} + {b}^{2} + {c}^{2} \\ \\ \sf \:$

Using the identity :-

• $\sf {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

$\sf {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ \\ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2ab - 2bc - 2ca \\ \\ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = ( {a + b + c)}^{2} - 2(ab + bc + ca) \\ \\ \sf \: now \: putting \: the \: value \: of \: ( {a + b + c)}^{2} and \: (ab + bc + ca) \longrightarrow \\ \\ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = ( {9)}^{2} - 2 \times (23) \\ \\ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = 81 - 46 \\ \\ </strong><strong>\</strong><strong>h</strong><strong>u</strong><strong>g</strong><strong>e</strong><strong> \boxed{ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = 35}$

Answer 2

$\huge\underline\bold{Answer}$

Given :-

● a + b + c = 9

● ab + bc + ca = 23

To find :-

☆ a² + b² + c² = ?

Solution :-

★

( a + b + c ) ² = a² + b² + c² + 2 ( ab + bc + ca)

=> ( 9 )² = a² + b² + c² + 2( 23 )

=> 81 - 46 = a² + b² + c²

=> a² + b² + c² = 38