Question

If $A$ and $B$ are the roots of $x^2-3x+p=0$ and $C,D$ are the roots of $x^2-12x+q=0$, where $a,b,c,d$ form a $GP$ then prove that $q+p:q-p=17:15$.

Answer 1

Since $a,b,c,d$ are in GP

Let, $a=a,b=ar,c=a{r}^2,d=a{r}^3$ .......... (1)

The standard quadratic equation is $a{x}^2+bx+c=0$

Then Sum of roots = $-\frac{b}{a}$

and Product of roots = $\frac{c}{a}$

Since $a,b$ are roots of equation $x^2-3x+p=0$

Thus, $a+b=\frac{-(-3)}{1},a\times b=\frac{p}{1}$

From (1)

$a+b=a(1+r)=3,ab=a^{2}r=p$

Similarly, $c+d=a{r}^2(1+r)=12,c\times d=a^2{r}^5=q$

Thus $\frac{a+b}{c+d}=\frac{a(1+r)}{a{r}^2(1+r)}=\frac{3}{12}$

$\therefore r^2=4\Rightarrow r=\pm 2$

Now, $\frac{q+p}{q-p}=\frac{a^2{r}^5+a^{2}r}{a^2{r}^5-a^{2}r}=\frac{a^{2}r(r^4+1)}{a^{2}r(r^4-1)}$

$\Rightarrow \frac{r^4+1}{r^4-1}=\frac{2^4+1}{2^4-1}=\frac{17}{15}$