Question

If a+b+c=9 and ab+bc+ca=40find a²+b²+c²

Answer 1

Answer:

Step-by-step explanation:

for a²+b²+c²

(a+b+c)²=(9)²

a²+b²+c²+2ab+2bc+2ac=81

a²+b²+c²+40=81

a²+b²+c²=41

Answer 2

$(a + b + c) = 9 \\ = > { (a + b) + (c) )}^{2} = {9}^{2} \\ = > {(a + b)}^{2} + {c}^{2} + 2c(a + b) = 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2ac + 2bc = 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) = 81$

Putting the given values,

We get ,

${a}^{2} + {b}^{2} + {c}^{2} + 2 \times 40 = 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 81 - 80 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 1$