If x = 1/(√5 − 2) and y = 1/(√5 + 2), find (1) x + y and xy, and (2) x² + y².

Step-by-step solution

1. 1

   Rationalise both denominators first. Multiplying by the conjugate turns each denominator into $\left(\sqrt5\right)^{2}-2^{2}=1$:

   $x=\frac{1}{\sqrt5-2}\cdot\frac{\sqrt5+2}{\sqrt5+2}=\sqrt5+2,\qquad y=\frac{1}{\sqrt5+2}\cdot\frac{\sqrt5-2}{\sqrt5-2}=\sqrt5-2$

   The denominators both come out as exactly $1$, which is why the numbers are so clean.

2. 2

   (1) Compute the sum. The $\pm 2$ cancel:

   $x+y=\left(\sqrt5+2\right)+\left(\sqrt5-2\right)=2\sqrt5$

3. 3

   Compute the product. This is a difference of squares:

   $xy=\left(\sqrt5+2\right)\left(\sqrt5-2\right)=5-4=1$

   Note $x$ and $y$ are reciprocals of each other — visible directly from their original definitions.

4. 4

   (2) Use the symmetric identity rather than squaring each term.

   $x^{2}+y^{2}=(x+y)^{2}-2xy$

   This avoids expanding two surd squares and is the whole point of first finding $x+y$ and $xy$.

5. 5

   Substitute.

   $x^{2}+y^{2}=\left(2\sqrt5\right)^{2}-2(1)=20-2=18$

   $\boxed{x+y=2\sqrt5,\quad xy=1,\quad x^{2}+y^{2}=18}$

6. 6

   Check directly. $x=4.23607$ and $y=0.23607$: $x+y=4.47214=2\sqrt5$ ✓, $xy=1.00000$ ✓, and $x^{2}+y^{2}=17.9443+0.0557=18.000$ ✓. The identity $x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)$ would similarly give $\left(2\sqrt5\right)^{3}-3\left(2\sqrt5\right)=40\sqrt5-6\sqrt5=34\sqrt5$.