Question

The sum of two numbers is 8 and the sum of their squares is 32. What is the smaller number?

Answer 1

The smaller number is `3`

(The other number is therefore `5`)

Explanation:

We could use `x and y` to represent the two numbers, but it is possible to form an equation using only one variable.

If two numbers add up to 8, and we let one of them be `x`, the other one will be the difference, which is: `8-x`

The sum of their squares is `34`

`x^2 + color(red)((8-x)^2)= 34" "larrcolor(red)((a-b)^2 = a^2 -2ab +b^2)`

`x^2 + color(red)(64-16x+x^2)= 34`

`2x^2 -16x +64-34 =0`

`2x^2 -16x +30 =0" "larrdiv 2`

`x^2 -8x +15=0" "larr` find factors of 15 which add to 8.

`(x-3)(x-5)=0`

Setting factor equal to `0` gives:

`x = 3 and x=5`

Check:

`3+5 = 8`

`3^2 +5^2 = 9+25 = 34`

Answer 2

Smallest number is `3`.

Explanation:

Let `a` and `b` be the numbers concerned.

We are told that:

`a+b= 8 -> b=8-a` (A)

`a^2+b^2 = 34` (B)

(A) in (B):`-> a^2+(8-a)^2 = 34`

`a^2 + 64 -16a +a^2 =34`

`2a^2 -16a +30 = 0`

`a^2 -8a+15=0`

`(a-3)(a-5)=0 -> a = +3 or +5`

Hence the smallest `a=3`

`a=3` in (A) `-> b= 8-3 = 5`

Hence the two numbers are 3 and 5, where 3 is the smallest.