Question

If a + b + c = 18 and a2 + b2 + c2 = 122, then find the value of ab + bc + ca​

Answer 1

a + b + c = 18

a² + b² + c² = 122

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(18)² = 122 + 2(ab + bc + ca)

324 = 122 + 2(ab + bc + ca)

2(ab + bc + ca) = 324 - 122

2(ab + bc + ca) = 202

ab + bc + ca = 202 ÷ 2

ab + bc + ca = 101

Answer 2

Answer:

The required value of ab + bc + ca is 101

Step-by-step explanation:

It is given that,

a + b + c = 18

and, $a^{2} + b^{2} + c^{2} = 122$

We know that,

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

or, (18)² = 122 + 2 ( ab + bc + ca ) [ ∵ a + b + c = 18 & a² + b² + c² = 122 )

or, 324 = 122 + 2 ( ab + bc + ca )

or, 2 ( ab + bc + ca ) = 324 - 122

or, 2 ( ab + bc + ca ) = 202

or, ab + bc + ca        =  $\frac{202}{2}$

or, ab + bc + ca        = 101