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Question Number 97512 by bemath last updated on 08/Jun/20

$$\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{for}\mathrm{which}\mathrm{the}$$$$\mathrm{quadratic}\mathrm{equation}(1-2\mathrm{k}){\mathrm{x}}^2-6\mathrm{kx}-1=0$$$$\mathrm{and}{\mathrm{kx}}^2-\mathrm{x}+1=0\mathrm{have}\mathrm{atleast}$$$$\mathrm{one}\mathrm{roots}\mathrm{in}\mathrm{common}\mathrm{are}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by john santu last updated on 08/Jun/20

$$\mathrm{Let}\mathrm{common}\mathrm{root}\mathrm{be}\beta .$$$$(1-2\mathrm{k}){\beta }^2-6\mathrm{k}\beta -1=0$$$$\mathrm{k}{\beta }^2-\beta +1=0$$$$\Rightarrow \frac{{\beta }^2}{-6\mathrm{k}-1}=\frac{\beta }{\mathrm{k}-1}=\frac{1}{{6\mathrm{k}}^2-(1-2\mathrm{k})}$$$$\Rightarrow {(\mathrm{k}-1)}^2=-(6\mathrm{k}+1)({6\mathrm{k}}^2+2\mathrm{k}-1)$$$${36\mathrm{k}}^3+{19\mathrm{k}}^2-6\mathrm{k}=0$$$$\mathrm{k}({36\mathrm{k}}^2+19\mathrm{k}-6)=0$$$$\mathrm{k}=0\mathrm{rejected}$$$${36\mathrm{k}}^2+19\mathrm{k}-6=0$$$$(4\mathrm{k}+3)(9\mathrm{k}-2)=0$$$$\{\begin{array}{l}\mathrm{k}=-\frac{3}{4}\\ \mathrm{k}=\frac{2}{9}\end{array}$$

Commented by bemath last updated on 08/Jun/20

$$\mathrm{thank}\mathrm{you}$$

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