Question

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

Answer 1

Given Quadratic equations are $x^2-3x+p=0$ $x^2-12x+q=0$ whose roots are $a,bandc,d$ respectively.

Let $a=A,$ b$=Ar$, $c$=$A{r}^2$, $d$= $A{r}^3$

Then $a,b,c,d$ form a$GP$ with a common ratio $r$

Now,$a+b=3$

$ab=p$

$c+d=12$

$cd=q$

$a+b=A(1+r)=3$

$A=\frac{3}{1+r}$

$c+d$ = $A{r}^2(1+r)$

From these two equations

$3r^2=12$

$r=\pm 2$

Let $r=2$

$A=\frac{3}{1+r}=1$

$a=A=1$

$b=Ar=2$

$c$= $A{r}^2=4$

$d$ = $A{r}^3=8$

$\frac{p+q}{p-q}=\frac{32+2}{32-2}=\frac{17}{15}$

Let $r=-2$

$A=\frac{3}{1-r}=-3$

$a=-3$

$b=Ar=6$

$c=A{r}^2=-12$

$d=A{r}^3=24$

$\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}$

Hence, Proved