# If the sum of two numbers is 4 and their product is 3, then what is the sum of their squares?

Feb 26, 2017

$10$

#### Explanation:

The two numbers are $1$ and $3$, being the two zeros of:

${x}^{2} - 4 x + 3 = \left(x - 1\right) \left(x - 3\right)$

In general we find:

$\left(x - a\right) \left(x - b\right) = {x}^{2} - \left(a + b\right) x + a b$

Notice that the coefficient of the middle term is $- \left(a + b\right)$ and the constant term is $a b$.

Given that the two numbers are $1$ and $3$, the sum of their squares is:

${1}^{2} + {3}^{2} = 1 + 9 = 10$

Feb 26, 2017

10

#### Explanation:

First, let's call the two numbers $m$ and $n$.

We can then write:

$m + n = 4$
$m n = 3$

We can solve the first equation for $m$:

$m + n - \textcolor{red}{n} = 4 - \textcolor{red}{n}$

$m + 0 = 4 - n$

$m = 4 - n$

Next, we can substitute $4 - n$ for $m$ in the second equation and solve for $n$:

$\left(4 - n\right) n = 3$

$4 n - {n}^{2} - 3 = 0$

${n}^{2} - 4 n + 3 = 0$

$\left(n - 1\right) \left(n - 3\right) = 0$

Solution 1)
$n - 1 = 0$

$n = 1$

Solution 2)
$n - 3 = 0$

$n = 3$

Substituting these back into the solution to the first equation gives:

Solution 1)

$m = 4 - 1$

$m = 3$

Solution 2)

$m = 4 - 3$

$m = 1$

The two numbers therefore are $3$ and $1$.

The sum of their squares is therefore:

${3}^{2} + {1}^{2} = 9 + 1 = 10$

Mar 30, 2018

$10$

#### Explanation:

Calling the two numbers $x$ and $y$, we are given:

$\left\{\begin{matrix}x + y = 4 \\ x y = 3\end{matrix}\right.$

and we find:

${x}^{2} + {y}^{2} = {x}^{2} + 2 x y + {y}^{2} - 2 x y$

$\textcolor{w h i t e}{{x}^{2} + {y}^{2}} = {\left(x + y\right)}^{2} - 2 x y$

$\textcolor{w h i t e}{{x}^{2} + {y}^{2}} = {4}^{2} - 2 \left(3\right)$

$\textcolor{w h i t e}{{x}^{2} + {y}^{2}} = 16 - 6$

$\textcolor{w h i t e}{{x}^{2} + {y}^{2}} = 10$