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Q.
If $p^{th}, q^{th}$ and $r^{th}$ terms of a G.P. be 27, 8, and 12 respectively, then the equation $px^{2}+2qx-2r=0$ has
Complex Numbers and Quadratic Equations
Solution:
According to problem
$27 = AR^{p-1}\, \dots(i)$
$8=AR^{q-1}\, \dots(ii)$
$12=AR^{r-1}\, \dots (iii)$
$\frac{27}{8} = R^{p-q} \, \dots(iv)$
[Using (i) & (ii)]
$\frac{12}{8} = R^{r-q}\,\dots(v)$
[Using (ii) & (iii)]
Now from (iv) and (v), we get
$\left(\frac{3}{2}\right)^{3}=R^{p-q}$ and $\left(\frac{3}{2}\right)^{1}=R^{r-q}$
$\left(\left(\frac{3}{2}\right)^{1}\right)^{3}=\left(R^{r-q}\right)^{3}$
$\Rightarrow R^{p-q}=R^{3r-3q}$
$\Rightarrow \left(p-q\right)log_{R} R =\left(3r-3q\right)log_{R} R $
$\Rightarrow p+2q-3r=0 \,\ldots\left(vi\right)$
Now, Assume $ f \left(x\right) =px^{2}+2qx-2r$
$\therefore f \left(0\right)=-2r $
and $f \left(1\right)=p+2q-2r$
and $f \left(1\right)=p+2q-2r$
$\therefore f \left(1\right)=r $ (From (vi))
Now, $f \left(0\right) f \left(1\right)=-2r^{2}<\,0$
$\therefore f \left(0\right)$ and $f \left(1\right) $ are having opposite signs, which means there exist only one root between $\left(0, 1\right)$
