Question

If \[a + b + c = 11\;\] and \[ab + bc + ac = 25\], then find the value of \[{a^2} + {b^2} + {c^2}\].

Answer 1

Hint: We will make use of the Identity \[{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}{\text{ }} = {\text{ }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} + {\text{ }}2ab + {\text{ }}2bc + {\text{ }}2ca\]
Substitute the given values into the identity and then we can find the required value.
When given an identity equation in certain variables, start by collecting like terms (terms of the same variable and degree) together. Doing this will usually pair terms one on one, thus making it easier to solve.
\[ \Rightarrow {\left( {a + b + c} \right)^2}{\text{ = }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} + {\text{ }}2\left( {ab + {\text{ }}bc + {\text{ }}ca} \right)\]
\[ \Rightarrow {a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}{\text{ = }}{\left( {a + b + c} \right)^2} - 2\left( {ab + {\text{ }}bc + {\text{ }}ca} \right)\]

Complete step-by-step answer:
Let's use the identity mentioned in the hint for finding the value of \[{a^2} + {b^2} + {c^2}\].
We have been given the value of \[a + b + c = 11\;\]and \[ab + bc + ac = 25\]
An equality true for every value of the variable in an expression is called identity. While factoring an algebraic expression using identities we use these to reach an irreducible form of the expression.
Identity which will be used:- \[{\left( {a + b + c} \right)^2}{\text{ = }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} + {\text{ }}2ab + {\text{ }}2bc + {\text{ }}2ca\]
By taking \[{a^2} + {b^2} + {c^2}\]on one side of the equation and other terms on other side of the equation, we get:
\[ \Rightarrow {a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}{\text{ = }}{\left( {a + b + c} \right)^2} - 2\left( {ab + {\text{ }}bc + {\text{ }}ca} \right)\]
We will now use the above condition to solve the problem:
Now substituting the value of \[a + b + c = 11\;\]and \[ab + bc + ac = 25\]in the above equation, we get:
\[ \Rightarrow {a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}{\text{ = }}{\left( {11} \right)^2} - 2\left( {25} \right)\]
Now opening both the brackets and solving the square, we get
\[ \Rightarrow {a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}{\text{ = 121}} - 50 = 71\]
\[ \Rightarrow {a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} = 71\]

So, the value of \[{a^2} + {b^2} + {c^2} = 71\].

Note: Proof for \[{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}{\text{ }} = {\text{ }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} + {\text{ }}2ab + {\text{ }}2bc + {\text{ }}2ca\]
Taking LHS of the identity and can also be written as:
$ \Rightarrow {\left( {a{\text{ }} + {\text{ }}b{\text{ }} + c} \right)^2} = {\text{ }}\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right){\text{ }}\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)$
Multiply as we do multiplication of trinomials and we get:
\[\begin{array}{*{20}{l}}
  { = {\text{ }}a\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right){\text{ }} + {\text{ }}b\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right){\text{ }} + {\text{ }}c\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)} \\
  { = {\text{ }}{a^2}{\text{ }} + {\text{ }}ab{\text{ }} + {\text{ }}ac{\text{ }} + {\text{ }}ab{\text{ }} + {\text{ }}{b^2}{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}ac{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}{c^2}}
\end{array}\]
Rearrange the terms and we get:
\[ = {\text{ }}{a^2}{\text{ }} + {\text{ }}{b^2}{\text{ }} + {\text{ }}{c^2}{\text{ }} + {\text{ }}ab{\text{ }} + {\text{ }}ab{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}ac{\text{ }} + {\text{ }}ac\]
Add like terms and we get:
\[ = {\text{ }}{a^2}{\text{ }} + {\text{ }}{b^2} + {\text{ }}{c^2}{\text{ }} + {\text{ }}2ab{\text{ }} + {\text{ }}2bc{\text{ }} + {\text{ }}2ca\]
Hence, in this way we obtain the identity i.e. \[{\left( {a + b + c} \right)^2}{\text{ = }}{a^2} + {\text{ }}{b^2} + {\text{ }}{c^2} + {\text{ }}2ab + {\text{ }}2bc + {\text{ }}2ca\]