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Question Number 83591 by jagoll last updated on 04/Mar/20

$${3\mathrm{x}}^2-\mathrm{x}+({\mathrm{t}}^2-4\mathrm{t}+3)=0$$$$\mathrm{has}\mathrm{a}\mathrm{roots}\sin \alpha \mathrm{and}\cos \alpha .$$$$\mathrm{find}\sqrt{{\mathrm{t}}^2-4\mathrm{t}+5}$$

Commented by jagoll last updated on 04/Mar/20

$$\sin \alpha +\cos \alpha =\frac{1}{3}$$$$\sin \alpha \times \cos \alpha =\frac{{\mathrm{t}}^2-4\mathrm{t}+3}{3}$$$$\Rightarrow 1+2\sin \alpha \cos \alpha =\frac{1}{9}$$$$\sin \alpha \cos \alpha =-\frac{4}{9}$$$$\frac{{\mathrm{t}}^2-4\mathrm{t}+3}{3}=-\frac{4}{9}$$$${\mathrm{t}}^2-4\mathrm{t}+3=-\frac{4}{3}$$$${\mathrm{t}}^2-4\mathrm{t}+5=\frac{2}{3}$$$$\sqrt{{\mathrm{t}}^2-4\mathrm{t}+5}=\frac{\sqrt{6}}{3}$$

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