Question

if (a+b+c) =15 and (ab+bc+ca) =47 find the value of (a^2+b^2+c^2)​

Answer 1

We know that,

(a+b+c)^2 = a^2+b^2+c^2 + 2ab+2bc+2ac .

Given, (a+b+c) = 15 and (ab+bc+ac) = 47

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

or, (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)

By putting the respected values, we get

(15)^2 = a^2 + b^2 + c^2 + 2(47)

or, 225 - 94 = a^2 + b^2 + c^2

or, 131 = a^2 + b^2 + c^2

The value of a^2 + b^2 + c^2 is 131.

Answer 2

Answer: 131

Step-by-step explanation:

Given that:

(a+b+c) = 15

(ab+bc+ca) = 47

To find : a^2+b^2+c^2

we know that (a+b+c) = 15

squaring both sides we get,

(a+b+c)^2 = (15)^2

=>(a+b+c) x (a+b+c) = 225

=>a^2+ab+ac + ab+b^2+bc + ac+bc+c^2 = 225

=> a^2+b^2+c^2 + 2(ab+bc+ca) = 225

=> a^2+b^2+c^2 + 2(47) = 225              (since ab+bc+ca = 47)

=>a^2+b^2+c^2 + 94 =225

=>a^2+b^2+c^2 = 225-94

Therefore, a^2+b^2+c^2 = 131 .

Hope it helps you.