vmatrix x+1 x x x x+ x x x x+ ^2 vmatrix=(9)/(8)(103x+81), then ,( )/(3) are the roots of …

Question

$\begin{vmatrix} x+1 \quad  x \quad  x \ x \quad  x+\lambda  \quad  x \ x \quad  x \quad  x+{\lambda }^{2} \end{vmatrix}=\frac{9}{8}(103x+81)$, then $\lambda ,\frac{\lambda }{3}$ are the roots of the equation :

Accepted Answer

$\begin{vmatrix} x+1 \quad  x \quad  x \ x \quad  x+\lambda  \quad  x \ x \quad  x \quad  x+{\lambda }^{2} \end{vmatrix}=\frac{9}{8}(103x+81)$

Put $x=0$

$\begin{gathered} |\begin{array}{ccc}1 \quad  0 \quad  0 \ 0 \quad  \lambda  \quad  0 \ 0 \quad  0 \quad  {\lambda }^{2}\end{array}|=\frac{9}{8}	imes 81 \ {\lambda }^{3}=\frac{{3}^{6}}{{2}^{3}} \end{gathered}$

$\begin{gathered} \Rightarrow \lambda =\frac{9}{2} \ \Rightarrow \frac{\lambda }{3}=\frac{3}{2} \end{gathered}$

$\begin{gathered} {x}^{2}-(\frac{9}{2}+\frac{3}{2})x+\frac{9}{2}	imes \frac{3}{2}=0 \ \Rightarrow 4{x}^{2}-24x+27=0 \end{gathered}$

Solutions · 1