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Question: If \[a + b + c = 12\] and \[ab + bc + ca = 22\], find \({a^2} + {b^2} + {c^2}\)....

If a+b+c=12a + b + c = 12 and ab+bc+ca=22ab + bc + ca = 22, find a2+b2+c2{a^2} + {b^2} + {c^2}.

Explanation

Solution

The square of the sum of three or more terms can be determined by the formula of the determination of the square of sum of two terms.
The formula of (a+b+c)2={\left( {a + b + c} \right)^2} = a2+b2+c2{a^2} + {b^2} + {c^2} +2ab+2bc+2ca + 2ab + 2bc + 2ca
we will learn to expand the square of a trinomial (a + b + c).\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right).
Let (b + c) = x\left( {b{\text{ }} + {\text{ }}c} \right){\text{ }} = {\text{ }}x
Then (a + b + c)2  = (a + x)2  = a2  + 2ax + x2{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\; = {\text{ }}{\left( {a{\text{ }} + {\text{ }}x} \right)^2}\; = {\text{ }}{a^2}\; + {\text{ }}2ax{\text{ }} + {\text{ }}{x^2}
  = a2  + 2a (b + c) + (b + c)2\; = {\text{ }}{a^2}\; + {\text{ }}2a{\text{ }}\left( {b{\text{ }} + {\text{ }}c} \right){\text{ }} + {\text{ }}{\left( {b{\text{ }} + {\text{ }}c} \right)^2}
= a2  + 2ab + 2ac + (b2  + c2  + 2bc)= {\text{ }}{a^2}\; + {\text{ }}2ab{\text{ }} + {\text{ }}2ac{\text{ }} + {\text{ }}({b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2bc)
= a2  + b2  + c2  + 2ab + 2bc + 2ca= {\text{ }}{a^2}\; + {\text{ }}{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2ab{\text{ }} + {\text{ }}2bc{\text{ }} + {\text{ }}2ca
Therefore,
(a + b + c)2  = a2  + b2  + c2  + 2ab + 2bc + 2ca{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\; = {\text{ }}{a^2}\; + {\text{ }}{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2ab{\text{ }} + {\text{ }}2bc{\text{ }} + {\text{ }}2ca
From the question first we find the square of (a + b + c)\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)than use second equation ab+bc+caab + bc + catake twice of this equation so, this is value of last three variable in the formula left hand side value we also known we find it at very first place. Now put all the values in the formula and get the result.
∴\therefore On substituting values, we get the required result.

Complete step by step solution:
Since a+b+c=12a + b + c = 12
∴  (a+b+c)2=144\therefore \,\,{\left( {a + b + c} \right)^2} = 144
And
ab+bc+ca=22ab + bc + ca = 22
2(ab+bc+ca)=442(ab + bc + ca) = 44
2ab+2bc+2ca=442ab + 2bc + 2ca = 44
∴\therefore On substituting these values, in the formula
(a + b + c)2  = a2  + b2  + c2  + 2ab + 2bc + 2ca{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\; = {\text{ }}{a^2}\; + {\text{ }}{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2ab{\text{ }} + {\text{ }}2bc{\text{ }} + {\text{ }}2ca
For unknown part take it to left hand side and in the right hand side we put those value we have, also remember when we change the variable from left hand side to right hand side or right hand side to left hand side the sign of the variable will change + will convert into – and – will convert into +.
a2+b2+c2{a^2} + {b^2} + {c^2} =(a+b+c)2−2(ab+bc+ca) = {\left( {a + b + c} \right)^2} - 2(ab + bc + ca)
{a^2} + {b^2} + {c^2}$$$ = 144 - 44$$ {a^2} + {b^2} + {c^2}$=100 = 100

Note: We can also derive this formula using (a+b)2{\left( {a + b} \right)^2} and letting (b+c)=x(b + c) = x
∴\therefore (a+x)2{\left( {a + x} \right)^2}=a2+x2+2ax{a^2} + {x^2} + 2ax.
Then putting xx.
=a2+(b+c)2+2a(b+c)= {a^2} + {\left( {b + c} \right)^2} + 2a(b + c)
(a+b+c)2={\left( {a + b + c} \right)^2} = a2+b2+c2{a^2} + {b^2} + {c^2} +2ab+2bc+2ca + 2ab + 2bc + 2ca