Question

 If ${{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ = 16}}$ and${\text{ab + bc + ca = 10}}$, find the value of${\text{a + b + c}}$.

Answer 1

Hint: We know that ${\left( {{\text{a + b + c}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2} + 2\left( {{\text{ab + bc + ca}}} \right)$ .We can use this formula to solve the given question by putting the values of${{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}$ and${\text{ab + bc + ca}}$, then simplify it to get the answer by removing the square-root.

Step-by-Ste Solution:

Given, ${{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ = 16}}$ and${\text{ab + bc + ca = 10}}$. We have to find ${\text{a + b + c}}$.Now we know the formula that ${\left( {{\text{a + b + c}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2} + 2\left( {{\text{ab + bc + ca}}} \right)$.So on putting the given values in this formula, we get-
$
   \Rightarrow {\left( {{\text{a + b + c}}} \right)^2} = 16 + 2\left( {{\text{10}}} \right) \\
   \Rightarrow {\left( {{\text{a + b + c}}} \right)^2} = 16 + 20 = 36 \\
 $
Now, on removing the square-root we get,
$ \Rightarrow {\text{a + b + c}} = \sqrt {36} = \pm 6$
Hence the answer is ${\text{a + b + c}} = 6$

Note: Here we are only taking positive value as the answer because every non-negative real number has a unique non-negative square-root, called the principal square-root.36, and 6 are non-negative in the question.