24\begin{bmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{bmatrix}=0 \ =>\alpha_1.\beta_2-\alpha_2\beta_1 =0 => \frac{\alpha_1}{\alpha_2}=\frac{\beta_1}{\beta_2}
if \alpha _{1},\alpha _{2} are the roots of equation:
ax2+bx+c=0
and\beta _{1},\beta _{2} be roots of equation :
px2+qx+r=0.if
\alpha _{1}y+\alpha_{2}z=0
\beta_{1}y+\beta_{2}z=0
have a non trivial(nonzero solutions) then prove that:
\frac{a}{p},\frac{b}{q}and ,\frac{c}{r}
are in g.p
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3 Answers
2424Apply C and D after this \frac{\alpha_1+\alpha_2}{\alpha_1-\alpha_2}=\frac{\beta_1+\beta_2}{\beta_1-\beta_2}
putting values and some reaagarnagemnts u get \frac{b^2-4ac^2}{b^2}=\frac{q^2-4pr}{q^2}
"*4ac above"
=> 1- \frac{4ac}{b^2}=1-\frac{4pr}{q^2}=>4ac/b^2=4pr/q^2=>ac/pr=b^2/q^2
=> GP
