Question

if a + b + c = 9 and ab + bc + ca = 40 , find
$a ^{2} + b ^{2} + c^{2}$
​

Answer 1

Answer:

$\large\bold\red{ {a}^{2} + {b}^{2} + {c}^{2} = 1}$

Step-by-step explanation:

It is being given that,

$a + b + c = 9$

and

$ab + bc + ca = 40$

To find :

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

As we already know the identity related to square of sum of three numbers, i.e.,

${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)$

So,

according to this,

we have simply,

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

Therefore,

putting the respective values,

we get,

$= > {(9)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + (2 \times 40) \\ \\ = > 81 = {a}^{2} + {b}^{2} + {c}^{2} + 8 0 \\ \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 81 - 80 \\ \\ = > \bold{ {a}^{2} + {b}^{2} + {c}^{2} = 1}$