# If x=2+sqrt(3) and y=2-sqrt(3) then what is x^2+y^2+xy ?

Sep 4, 2015

15

#### Explanation:

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

$= {\left(2 + \sqrt{3} + 2 - \sqrt{3}\right)}^{2} - \left[\left(2 + \sqrt{3}\right) \left(2 - \sqrt{3}\right)\right]$

$= 16 - \left(4 - 3\right) = 15$

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

Sep 4, 2015

$15$

#### Explanation:

If $x = 2 + \sqrt{3}$
$\textcolor{w h i t e}{\text{XXXXXXXX}} {x}^{2} = 4 + 4 \sqrt{3} + 3 = \textcolor{red}{7 + 4 \sqrt{3}}$

If $y = 2 - \sqrt{3}$
$\textcolor{w h i t e}{\text{XXXXXXXX}} {y}^{2} = 4 - 4 \sqrt{3} + 3 = \textcolor{red}{7 - 4 \sqrt{3}}$

and
$\textcolor{w h i t e}{\text{XXXXXXXX")xy = 4-3 color(white)("XXXXX}} = \textcolor{red}{1}$

Sum$= \textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXX}} = \textcolor{b l u e}{15}$

Sep 4, 2015

${x}^{2} + {y}^{2} + x y = 15$

#### Explanation:

Use the difference of squares identity:

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

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

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

So:

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

$= {\left(x + y\right)}^{2} - x y = {4}^{2} - 1 = 16 - 1 = 15$