$S$ and $G$ are roots of $x^2 −3x +1=0$. Find equation whose roots are $\frac{1}{S -2}$ and$\frac{1}{G-2}$

Question

If S,G are roots of $x^2 −3x +1=0$, then find equation whose roots are $\frac{1}{S -2}$ , $\frac{1}{G-2}$.

So , the way my sir solved it is that

He took x = $\frac{1}{S -2}$ , then he got S = $\frac{2x+1}{x}$.

I am confused with is that we were told roots as $\frac{1}{S -2}$ and not only S for 2nd equation.

Why not just put $\frac{1}{S -2}$ in x and then solve ?

Answer 1

Note $SG=1,\>S+G=3$. Then

$$\frac1{S-2}\frac1{G-2}=\frac1{SG-2(S+G)+4}=-1 $$ $$\frac1{S-2}+\frac1{G-2}=\frac{S+G-4}{SG-2(S+G)+4}=1 $$ Thus, the equation with roots $\frac1{S-2}$ and $\frac1{G-2}$ is $$x^2-x-1=0$$

Answer 2

The substitution method your teacher is referring to goes as follows:

Let $y=\frac{1}{x-2}$ where $x=S, G$ are the roots of the given equation $x^2-3x+1=0$

Rearranging gives $x=\frac{1+2y}{y}$

Substituting this into the equation satisfied by $x$ gives $$(\frac{1+2y}{y})^2-3(\frac{1+2y}{y})+1=0$$

This simplifies to become $$y^2-y-1=0$$ This is the same as the answer obtained by Quanto using Vieta's formulas.