Question

If a + b + c = 8 and ab + bc + ca = 19, then find a2+b2+c2



#No - Spams!

Answer 1

$\huge{\boxed{\blue{\it{Answer:-}}}}$

Given:-

a+b+c=8 -----(1)

ab+bc+ca=19---(2)

By algebraic identity:-

a²+b²+c² = (a+b+c)²-2(ab+bc+ca)

= 8²-2×19 [From (1) & (2)]

= 64 - 38

$\large{\boxed{\tt{ = 26}}}$

Therefore,

a^2 + b^ 2 +c^ 2 =26

Answer 2

$\huge\bigstar\underline\mathfrak\green{Correct\:question}$

If a + b + c = 8 and ab + bc + ca = 19, then find a²+b²+c²

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$\huge\bigstar\underline\mathfrak\green{Answer}$

The value of a² + b² + c² = 26

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$\huge\bigstar\underline\mathfrak\green{Explanation}$

Given : a + b + c = 8 and ab + bc + ca = 19

To find : The value of a² + b² + c²

Solution : We know that,

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

After taking 2 common,

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

Now, put the given values in the above identity.

=> ( 8 )² = a² + b² + c² + 2(19)

=> 64 = a² + b² + c² + 38

=> 64 - 38 = a² + b² + c²

=> a² + b² + c² = 26 ( required answer )

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