Question

if a+b+c=9 and ab+bc+ca=40 then find the value of a²+b²+c²?

Answer 1

Heya ✋

Let see your answer !!!!

a + b + c = 9

ab + bc + ca = 40

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

a + b + c = 9

On squaring both sides

$(a + b + c) ^{2} = 9 ^{2} \\ = > a ^{2} + b ^{2} + c ^{2} + 2ab + 2bc \\ + 2ca = 81 \\ = > a ^{2} + b ^{2} + c ^{2} + 2(ab + bc \\ + ca) = 81 \\ = > a ^{2} + b ^{2} + c ^{2} + 2 \times 40 = \\ 81 \\ = > a ^{2} + b ^{2} + c ^{2} + 80 = 81 \\ = > a ^{2} + b ^{2} + c ^{2} = 81 - 80 \\ = > a ^{2} + b ^{2} + c ^{2} = 1$




Thanks :))))

Answer 2

We Know that....

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

so by putting the values...

9^2=a^2+b^2+c^2+40

or ...

a^2+b^2+c^2=81-40

=41