Question

The sum of two numbers is 25, the sum of their squares is 313. What are the numbers?

Answer 1

`13` and `12`

Explanation:

If `x+y=25` then `y=25-x`.

Substitute:
`x^2+y^2=313`
`x^2+(25-x)^2=313`
`2x^2-50x+312=0`
`x^2-25x+156=0`
`(x-13)(x-12)=0`

`x=13,y=12`
`x=12,y=13`

Answer 2

`12` and `13`.

Explanation:

Suppose the two numbers are `a` and `b`

"The sum of two numbers is 25" gives us:

`a+b=25`

"sum of their squares is 313" gives us:

`a^2+b^2=313`

From the first equation we have:

`b=25-a`

Substituting for `b` into the second we get:

`a^2+(25-a)^2=313`

`:. a^2+625-50a+a^2=313`
`:. 2a^2-50a+312=0`
`:. a^2-25a+156=0`
`:. (a-12)(a-13)=0`
`:. a=12,13`

We now use the second equation to find the value of `b` corresponding to each solution.

`a=12 => b=25 - 12 = 13`
`a=13 => b=25 - 13 = 12`

So there is only one solution which is that the two numbers are `12` and `13`.

Answer 3

The numbers are `12` and `13`

Explanation:

Let te numbers be `x` and `y`

therefore we have `x+y=25` ............(1)

which gives us `x^2+y^2+2xy=625` ............(2)

and we also have `x^2+y^2=313` ............(3)

Subtracting (3) from (2), we get `2xy=312` ............(4)

and (4) from (3) we get

`x^2+y^2-2xy=1` ............(5)

i.e. `x-y=1` ............(6)

Slolving for `x` and `y` from (1) and (6), we get

`x=13` and `y=12`

Note:we can also have `x-y=-1`, which gives `x=12` and `y=13`