Question

. If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 + c 2

Answer 1

Answer :

Given :

a + b + c = 9

ab + bc + ca = 26

Required to find :

${a}^{2} + {b}^{2} + {c}^{2}$

Solution:

The given sum is very easy .

Remember , whenever you see this type of questions try solving this using the first statement.

Because, In most cases we can solve this type of polynomial questions using the first statement.

So, let's coming to the question

It is given that a + b + c = 9

so using this we are going to solve this question .

Since, the question contains square let's do squaring on both sides

That is

a + b + c = 9

=> squaring on both sides

Then ,

(a + b + c)^2 = (9)^2

As we know that (a + b + c)^2 is an identity let's expand it

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

so,

${a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \: = 81$

Here we can see that 2 is common in the identity So , take that as a common .

Therefore,

${a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 81$

In the above he clearly mentioned that ,

ab + bc + ca = 26

so , Substitute this value in place of ab + bc + ca

Then we get

${a}^{2} + { b}^{2} + {c}^{2} + 2(26) = 81$

Now, By transposing the constants either side and solving further we get

${a}^{2} + {b}^{2} + {c}^{2} + 52 = 81 \\ {a}^{2} + {b}^{2} + {c}^{2} = 81 - 52 \\ {a}^{2} + {b}^{2} + {c}^{2} = 29$

Therefore,

we get

${a}^{2} + {b}^{2} + {c}^{2} = 29$