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

Mar 23, 2017

The smaller number is $3$

(The other number is therefore $5$)

#### Explanation:

We could use $x \mathmr{and} y$ 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 $34$

${x}^{2} + \textcolor{red}{{\left(8 - x\right)}^{2}} = 34 \text{ } \leftarrow \textcolor{red}{{\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}}$

${x}^{2} + \textcolor{red}{64 - 16 x + {x}^{2}} = 34$

$2 {x}^{2} - 16 x + 64 - 34 = 0$

$2 {x}^{2} - 16 x + 30 = 0 \text{ } \leftarrow \div 2$

${x}^{2} - 8 x + 15 = 0 \text{ } \leftarrow$ find factors of 15 which add to 8.

$\left(x - 3\right) \left(x - 5\right) = 0$

Setting factor equal to $0$ gives:

$x = 3 \mathmr{and} x = 5$

Check:

$3 + 5 = 8$

${3}^{2} + {5}^{2} = 9 + 25 = 34$

Mar 23, 2017

Smallest number is $3$.

#### Explanation:

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

We are told that:

$a + b = 8 \to b = 8 - a$ (A)

${a}^{2} + {b}^{2} = 34$ (B)

(A) in (B):$\to {a}^{2} + {\left(8 - a\right)}^{2} = 34$

${a}^{2} + 64 - 16 a + {a}^{2} = 34$

$2 {a}^{2} - 16 a + 30 = 0$

${a}^{2} - 8 a + 15 = 0$

$\left(a - 3\right) \left(a - 5\right) = 0 \to a = + 3 \mathmr{and} + 5$

Hence the smallest $a = 3$

$a = 3$ in (A) $\to b = 8 - 3 = 5$

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