Question

The sum of the squares of three numbers is 83, while sum of their products taken two at a time is 71. Their sum will be​

Answer 1

Answer:

15

Step-by-step explanation:

Let three numbers be x, y and z.

A/Q,

The sum of the squares of three numbers = 83

       $x^{2} +y^{2} +z^{2} =83$

The sum of their products taken two at a time = 71

       $xy + yz+zx = 71$

As we know,

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

$(x+y+z)^{2} =x^{2} +y^{2} +z^{2} +2(xy + yz + zx)\\=> (x+y+z)^{2} =83 + 2(71)\\=>(x+y+z)^{2} = 83 + 142\\=> (x+y+z)^{2} = 225\\=> (x+y+z) = \sqrt{225} \\x+y+z = 15$

Thus, the sum of these numbers = 15

Answer 2

Answer:

Let the numbers be a, b and c

Then, a2 + b2 + c2 = 83

and (ab + bc + ca) = 71

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

= 83 + 2 x 71 = 225

⇒ (a + b + c) = ✓225