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

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Answer 1 · 2017-03-23T11:11:25

The smaller number is $3$

(The other number is therefore $5$ )

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 · 2017-03-23T11:45:08

Smallest number is $3$ .

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.