If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p {\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q.

Question

If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p{\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q.

So I take the second term $4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2}) (r)$ , where r is the common ratio.

$4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2})( 2{\sqrt3} + {\sqrt2 })$

And found that the common ratio, r = $2{\sqrt3} + {\sqrt2 }$

To find the third term, I multiplied the second term with the common ratio.

$(4-{\sqrt 6})( 2{\sqrt3} + {\sqrt2 })= p{\sqrt 3} - q {\sqrt 2}$

$8{\sqrt 3} + 4{\sqrt2} - 6 {\sqrt 2} - 2{\sqrt 6} = p{\sqrt 3} - q {\sqrt 2}$

I am unable to proceed beyond this step.

Answer 1

Note

$$(4- {\sqrt 6})^2=({\sqrt 3} - {\sqrt 2})( p {\sqrt 3} - q {\sqrt 2})$$

or,

$$22-8\sqrt6 = 3p +2q -(p+q)\sqrt6$$

Therefore,

$$22= 3p +2q,\>\>\>\>\>p+q=8$$

Solve to obtain $p=6$ and $q=2$.

Answer 2

If it means that $p$ and $q$ are rationals, so use the following:$$(\sqrt3-\sqrt2)(p\sqrt3-q\sqrt2)=(4-\sqrt6)^2$$ or $$3p+2q-(p+q)\sqrt6=22-8\sqrt6.$$ I got $$(p,q)=(6,2).$$

Answer 3

You've made a computation error, perhaps? $$(4-\sqrt6)(2\sqrt3+\sqrt2)=8\sqrt3+4\sqrt2-6\sqrt2-2\sqrt3=6\sqrt3-2\sqrt2$$