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Question Number 78888 by zainal tanjung last updated on 21/Jan/20

$$\mathrm{The}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$(x-a)(x-b)={abx}^2\mathrm{are}\mathrm{always}$$

Commented by mr W last updated on 21/Jan/20

$$...arealwayswhat?$$

Answered by behi83417@gmail.com last updated on 21/Jan/20

$$(1-\mathrm{ab}){\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}=0$$$$△={(\mathrm{a}+\mathrm{b})}^2-4\mathrm{ab}(1-\mathrm{ab})=$$$$={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}-4\mathrm{ab}+{4\mathrm{a}}^2{\mathrm{b}}^2=$$$$={(\mathrm{a}-\mathrm{b})}^2+{4\mathrm{a}}^2{\mathrm{b}}^2>0[\mathrm{\forall }\mathrm{a},\mathrm{b}\in \mathrm{R}]$$$$\Rightarrow \mathrm{there}\mathrm{is}\mathrm{alweys}\mathrm{two}\mathrm{real}\mathrm{roots}.$$

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