If x = (sqrt3 + 1)/(sqrt3 - 1) and y = (sqrt3 - 1)/(sqrt3 + 1), find x + y, xy, x^2 + y^2, x^3 + y^3 and 4x^2 + 7xy + 4y^2.

Step-by-step solution

1. 1

   Rationalise each denominator first — every later step becomes trivial once the radicals are out of the bottom. Multiply numerator and denominator by the conjugate of the denominator:

   $x=\frac{\sqrt3+1}{\sqrt3-1}\cdot\frac{\sqrt3+1}{\sqrt3+1}=\frac{(\sqrt3+1)^2}{3-1}=\frac{4+2\sqrt3}{2}=2+\sqrt3$

   $y=\frac{\sqrt3-1}{\sqrt3+1}\cdot\frac{\sqrt3-1}{\sqrt3-1}=\frac{(\sqrt3-1)^2}{3-1}=\frac{4-2\sqrt3}{2}=2-\sqrt3$

   Notice $x$ and $y$ came out as conjugates, and $y=1/x$: the two fractions were reciprocals from the start.

2. 2

   Compute the two basic symmetric quantities. Every remaining expression can be built from these two, so they are the only genuine computations in the problem:

   $x+y=(2+\sqrt3)+(2-\sqrt3)=4$

   $xy=(2+\sqrt3)(2-\sqrt3)=4-3=1$

   The radicals cancel in both — that is the whole reason to work with $x+y$ and $xy$ rather than with $x$ and $y$ separately.

3. 3

   Get $x^2+y^2$ from the square of the sum. Expanding $(x+y)^2=x^2+2xy+y^2$ and rearranging,

   $x^2+y^2=(x+y)^2-2xy=4^2-2(1)=14.$

4. 4

   Get $x^3+y^3$ from the cube of the sum. Since $(x+y)^3=x^3+y^3+3xy(x+y)$,

   $x^3+y^3=(x+y)^3-3xy(x+y)=4^3-3(1)(4)=64-12=52.$

   (The factorisation $x^3+y^3=(x+y)(x^2-xy+y^2)=4(14-1)=52$ gives the same value, a useful cross-check.)

5. 5

   Regroup the last expression so it only uses known quantities.

   $4x^2+7xy+4y^2=4(x^2+y^2)+7xy=4(14)+7(1)=56+7=63.$

   The trick is to split the middle coefficient: $7xy=8xy-xy$ would also work, but pulling out $4(x^2+y^2)$ directly is shorter.

6. 6

   Check numerically. $x=2+\sqrt3\approx3.7320508$ and $y=2-\sqrt3\approx0.2679492$. Then $x+y\approx4.000000$, $xy\approx1.000000$, $x^3+y^3\approx51.999999$ and $4x^2+7xy+4y^2\approx63.000000$, matching the exact values.