Q&A - Ask Doubts and Get Answers
Q

If a and b are the roots of 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.

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

Answers (1)
Views

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

Then, a+b=3\, \, \, \, and\, \, \, \, ab=p......................1

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

c+d=12\, \, \, \, and\, \, \, \, cd=q......................2

Given: a, b, c, d form a G.P

Let, a=x,b=xr,c=xr^2,d=xr^3

From 1 and 2, we get

 x+xr=3                      and           xr^2+xr^3=12

\Rightarrow x(1+r)=3                              xr^2(1+r)=12

On dividing them, 

        \frac{xr^2(1+r)}{x(1+r)}=\frac{12}{3}

\Rightarrow r^2=4

\Rightarrow r=\pm 2

When , r=2 ,     

                    x=\frac{3}{1+2}=1

When , r=-2,

                    x=\frac{3}{1-2}=-3

CASE (1) when r=2 and x=1,

ab=x^2r=2\, \, \, and \, \, \, \, cd=x^2r^5=32

\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}

i.e. (q + p) : (q – p) = 17:15.

CASE (2) when r=-2 and x=-3,

ab=x^2r=-18\, \, \, and \, \, \, \, cd=x^2r^5=-288

\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-305}{-270}=\frac{17}{15}

i.e. (q + p) : (q – p) = 17:15.

Exams
Articles
Questions