Quadratic expression in $x$ with roots $\frac gh$ and $-\frac hg$ What quadratic expression in $x$ has roots $$\frac{g}{h}\qquad\text{and}\qquad-\frac{h}{g}?$$ I know that this can be factored as $$\left ( x-\frac{g}{h} \right )\left ( x+\frac{h}{g} \right )=0$$ But I'm really confused with the choices. The choices are as follows:
$(a)\hspace{5 mm}g^2 h^2 x^2 -\frac{g^2}{h^2}$
$(b)\hspace{5 mm}hgx^2+(g^2-h^2)x-hg$
$(c)\hspace{5 mm}hgx^2+(h^2-g^2)x+hg$
$(d)\hspace{5 mm}hgx^2+(h^2-g^2)x-hg$
Is the answer (d)?
 A: Given:
$$\boxed{\left ( x-\frac{g}{h} \right )\left ( x+\frac{h}{g} \right )=0}$$
Start by expanding the factors using the distributive property, sometimes referred to as FOIL: 
$$x^2+\dfrac{xh}{g}-\dfrac{xg}{h}-\dfrac{hg}{hg}=0$$
We shall next multiply both sides of the equation by the common denominator $hg$. The purpose of this is to remove the denominator making the equation easier to manage:
$$hg\bigg(x^2+\dfrac{xh}{g}-\dfrac{xg}{h}-\dfrac{hg}{hg}=0\bigg)$$
$$hg\bigg(hgx^2+xh^2-{xg^2}-{hg}=0\bigg)$$
As we can see, the equation is quite different from the original now and free of fractions:
$$hgx^2+x(xh^2-xg^2)-{hg}=0$$
Factoring out $x$ from $xh^2-xg^2$
$$\boxed{hgx^2+x(h^2-g^2)-{hg}=0}$$
Which gives us our final answer of: D
A: $\displaystyle\begin{eqnarray}{\bf Hint}\!: &&0 &\, =& \ \ \left (x-\frac{g}{h} \right )\ \ \left ( x+\frac{h}{g} \right ) \\
\stackrel{\large\times\,\color{#0a0}h\color{#c00}g}\Rightarrow && 0 &\,=& \color{#0a0}h\left ( x-\frac{g}{h} \right ) \color{#c00}g\left ( x+\frac{h}{g} \right )\\
            &&   &\,=&\ \  (h\,x -g)\ \ \ \ (g\,x + h)
\end{eqnarray}$
