Question

a2+b2+c2=74 and ab+bc+ca=61 so find a+b+c​

Answer 1

Step-by-step explanation:

اللہ آپ کا بھلا کرے..........!

Answer 2

The value of $(a+b+c)^{2}$= 196

Step-by-step explanation:

We have,

$a^2+b^2+c^2$ = 74 and ab + bc + ca  = 61

To find, the value of $(a+b+c)^{2}$= ?

We know that,

The algebraic identity,

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

Put $a^2+b^2+c^2$ = 74 and ab + bc + ca  = 61, we get

∴ $(a+b+c)^{2}$ = 74 + 2(61)  

⇒ $(a+b+c)^2$ = 74 + 122 = 196

∴ $(a+b+c)^2$ = 196

Thus, the value of $(a+b+c)^{2}$= 196

You can refer:

https://brainly.in/question/14425960