Question

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

Answer 1

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

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

$\Rightarrow a+b=3$ and $ab=p$ ....(1)

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

$c+d=\frac{-(-12)}{1}$ and $cd=q$

$\Rightarrow c+d=12$ and $cd=q$ ....(2)

Also, given $a,b,c,d$ are in G.P.

So, let $a=x,b=rx,c=r^{2}x,d=r^{3}x$ ....(3)

Put the values from eq(3) in eq(1), we get

$x+rx=3$ .....(4)

Put the values from eq(3) in eq(2), we get

$r^{2}x+r^{3}x=12$

$r^2(x+rx$=12$

$\Rightarrow r^2\left(3\right)=12$ (by (4))

$\Rightarrow r^2=4$

$\Rightarrow r=\pm 2$

Using this in eqn (4),

When $r=2$,

$x+2x=3$

$\Rightarrow x=1$

When $r=-2$

$\Rightarrow x-2x=3$

$\Rightarrow x=-3$

Since, by eq (1) and (2), we also have

$ab=p$ and $cd=q$

$r{x}^2=p$ and $r^5{x}^2=q$ ...(5)

Case I: When $r=2,x=1$

$2(1{)}^2=p$ and $2^5(1{)}^2=q$

$\Rightarrow p=2$ and $q=32$

$\Rightarrow \frac{q+p}{q-p}=\frac{34}{30}=\frac{17}{15}$

Case II: When $r=-2,x=-3$

$-2(-3{)}^2=p$ and $(-2{)}^5(-3{)}^2=-288$ (by (5))

$\Rightarrow p=-18$ and $q=-288$

$\Rightarrow \frac{q+p}{q-p}=\frac{-306}{-270}=\frac{17}{15}$

Hence, ratio of $q+p$ and $q-p$ is 17:15