Question

if a^2+b^2+c^2=16 and ab +bc+ca=10 find value of a+b+c​

Answer 1

Question:-

If a² + b² + c² = 16 and ab + bc + ca = 10, find the value of a + b + c

$\huge\bf\underline{Solution}$

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

Putting the values.

=> (a + b + c)² = 16 + 2 (10)

=> (a + b + c)² = 16 + 20

=> (a + b + c)² = 36

=> (a + b + c) = √36

=> (a + b + c) = 6

Hence, the value of (a + b + c) is 6 respectively.

Identity used :-

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

Some Important identities :-

$\bullet$(a + b)² = a² + b² + 2ab

$\bullet$(a - b)² = a² + b² - 2ab

$\bullet$a² - b² = (a + b)(a - b)

$\bullet$a² + b² = (a + b)² - 2ab or (a - b)² + 2ab

$\bullet$(a + b)² - (a - b)² = 4ab

Answer 2

Answer :-

a + b + c = ± 6

Given:-

a² + b² + c² = 16

ab + bc + ca = 10

To find :-

Value of a + b + c.

Solution :-

For, solving this question we need a suitable identity I. e, :-

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

Now, put the above given value,

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

→$(a+b+c) ^2 = 16 + 2.10$

→$(a+b+c) ^2 = 16 + 20$

→$(a + b+c)^2 = 36$

→$(a + b+c) = \sqrt{36}$

→$(a+b+c) = \pm 6$

hence, the required value of :-

a + b + c = ± 6

Note :- Identities are important when u solve this type of question so, please learn it and try to understand by practice.