Question

Find a b + BC + CA if a + b + c = 29 and a² + b² + c² = 35

Answer 1

Correction in question :

Q. Find ( ab + bc + ca ) if ( a + b + c = 29 ) and ( a² + b² + c² = 35 ) .

To solve these types of problems you just need to concentrate on given things and what is to find.

Here we go,

Given,

⇒ ( a² + b² + c² ) = 35 ------- ( 1 )

⇒ ( a + b + c ) = 29 ------- ( 2 )

Squaring both sides,

⇒ ( a + b + c )² = 29²

⇒a² + b² + c² + 2ab + 2bc + 2ac = 841

⇒ ( a² + b² + c² ) + 2( ab + bc + ac ) = 841

Substitute the value of eq.( 1 ),

⇒ 35 + 2( ab + bc + ac ) = 841

⇒ 2( ab + bc + ca ) = 841 - 35

⇒ 2( ab + bc + ca ) = 806

⇒( ab + bc + ca ) = 806 ÷ 2

•°• ( ab + bc + ca ) = 403

Verification :

⇒ ( a + b + c ) = 29

Squaring both sides,

⇒ ( a + b + c )² = 29²

⇒ a² + b² + c² + 2ab + 2bc + 2ac = 841

⇒ a² + b² + c² + 2( ab + bc + ca ) = 841

⇒ 35 + 2 ( 403 ) = 841

⇒ 35 + 806 = 841

•°• 841 = 841

Verified !!

Identity used :

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

The final answer is 403.

Answer 2

✴✴$\huge \bf{HEY \: FRIENDS!!}$✴✴

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✴✴$\huge \bf \underline{Here \: is \: your \: answer↓}$⬇⏬⤵

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▶⏩ It is given that:-)

$\boxed{a + b + c = 29.}$
$\boxed{ {a}^{2} + {b}^{2} + {c}^{2} = 35.}$

▶⏩ Now,

↪➡$\bf{a + b + c = 29.}$

$\huge \boxed{squaring \: both \: side }$

↪➡$\bf{ {(a + b + c)}^{2} = {(29)}^{2}.}$

▶⏩using identity :-)

$\huge \boxed{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca).}$

↪➡$\bf{ {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 841.}$

↪➡$\bf{35 + 2(ab + bc + ca) = 841.}$

↪➡$\bf{2(ab + bc + ca) = 841 - 35.}$

↪➡$\bf{2(ab + bc + ca) = 806.}$

↪➡$\bf{ ab + bc + ca = \frac{806}{2}.}$

$\huge \bf{ ab + bc + ca = 403.}$

✅✅ Hence, it is finded. ✔✔

✴✴$\huge \boxed{THANKS}$✴✴

☺☺☺$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$✌✌✌.