Question

If a and b are the roots $x^2-3x+p=0$ and c, d are roots of $x^2-12x+q=0$, where a, b, c, d form a G.P. Prove that $(q+p):(q-p)=17:15$.

Answer 1

a, b, c, d are in A.P.

Let $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=k$

$\therefore \frac{b}{a}=k\Rightarrow b=ak$

$\frac{c}{b}=k\Rightarrow c=bk=\left(ak\right)k=a{k}^2$

$\frac{d}{c}=k\Rightarrow d=ck=\left(a{k}^2\right)k=a{k}^3$

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

$\therefore (a+b)=\frac{-(-3)}{1}$ and $ab=\frac{p}{1}$

$\Rightarrow a+ak=3$ and a, ak = p $[\because b=ak]$

$\Rightarrow a(1+k)=3\dots \dots \left(1\right)$

and $a^{2}k=p\dots \dots \left(2\right)$

Also, c and d are the roots of $x^2-12x+q=0$

$\therefore c+d=\frac{-(-12)}{1}$ and $d=\frac{2}{1}$

$\Rightarrow a{k}^2+a{k}^3=12$ and $a{k}^2.a{k}^3=q$

$\because c=a{k}^2$ and $d=a{k}^3$

$\Rightarrow a{k}^2(1+k)=12\dots \dots \left(3\right)$

and $a62k^5=q\dots \dots \left(4\right)$

Dividing (3) by (1),

$\frac{a{k}^2(1+k)}{a(1+k)}=\frac{12}{3}$

$\Rightarrow k^2=4$ or $k=\pm 2$

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

$=\frac{(k^4+1)}{(k^4-1)}=\frac{(\pm 2{)}^4+1}{(\pm 2{)}^4-1}$

$=\frac{16+1}{16-1}=\frac{17}{15}$

Thus, $(q+p):(q-p)=17:15$