Question

if the roots of a x³-13 x ²+ kx-27 =0 are in G.P. then k=
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Answer 1

$\textbf{Given:}$

$\text{Roots of x^3-13x^2+kx-27=0 are in G.P}$

$\textbf{To find:}$

$\text{The value of k}$

$\textbf{Solution:}$

$\text{Let the roots be \frac{a}{r},\,a,\,ar }$

$S_1=\dfrac{a}{r}+a+ar=13$

$S_2=\dfrac{a}{r}.a+a.ar+ar.\dfrac{a}{r}=k$

$S_3=\dfrac{a}{r}.a.ar=27$

$\implies\,a^3=27$

$\implies\bf\,a=3$

$S_1=\dfrac{3}{r}+3+3r=13$

$\dfrac{3+3r+3r^2}{r}=13$

$3+3r+3r^2=13r$

$3r^2-10r+3=0$

$3r^2-9r-r+3=0$

$3r(r-3)-1(r-3)=0$

$(3r-1)(r-3)=0$

$\implies\,r=3,\dfrac{1}{3}$

$\text{Then,}$

$k=\dfrac{a}{r}.a+a.ar+ar.\dfrac{a}{r}$

$k=\dfrac{3}{3}.3+3.3(3)+3(3).\dfrac{3}{3}$

$k=1.3+3.3(3)+3(3)$

$k=3+27+9$

$\implies\,k=39$

$\textbf{Answer:}$

$\textbf{The value of k is 39}$