Question

Let $a$ and $b$ be the roots of $x^2-3x+p=0$ and let $c$ and $d$ be the roots of $x^2-12x+q=0$, where $a,b,c$ form an increasing G.P. Then the ratio of $q+p:q-p$ is equal to

• A. $8:7$
• B. $11:10$
• C. $17:15$
• D. none of these

Answer 1

The correct option is C $17:15$

Given a,b,c,d are in G.P

$\Rightarrow \frac{b}{a}=\frac{c}{d}=\frac{d}{c}=k$ (say)

$\Rightarrow b=ak,c=bk,d=ckorb=ak,c=a{k}^2,d=a{k}^3......\left(1\right)$

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

$\Rightarrow a+b=3,ab=p$

$\Rightarrow a+ak=3,a.ak=p$ [By using (1)]

$\Rightarrow a(1+k)=\mathrm{3........}\left(2\right)$

and $a^{2}k=p.......\left(3\right)$

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

$\Rightarrow c+d=12,cd=q$

$\Rightarrow a{k}^2+a{k}^3=12,\left(a{k}^2\right)\left(a{k}^3\right)=q$ [By using (1)]

$\Rightarrow a{k}^2(1+k)=\mathrm{12....}\left(4\right)$

and $a^2{k}^5=q.....\left(5\right)$

Dividing (4) by (2), we get

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

$\Rightarrow k^2=\mathrm{4.....}\left(6\right)$

Consider $\frac{q}{p}=\frac{a^2{k}^5}{a^{2}k}=k^4=(k^2{)}^2=4^2=16$

$\Rightarrow \frac{q}{p}=16$

$\Rightarrow \frac{q}{p}=\frac{16}{1}$

Applying Componendo and dividendo, we get

$\frac{q+p}{q-p}=\frac{17}{15}$

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