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Question Number 77180 by jagoll last updated on 04/Jan/20

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

Answered by mr W last updated on 04/Jan/20

$$\sin \alpha +\cos \alpha =\frac{1}{3}...\left(i\right)$$$$\sin \alpha \cos \alpha =\frac{t^2-4t+3}{3}...\left(ii\right)$$$${\left(i\right)}^2:$$$$1+2\sin \alpha \cos \alpha =\frac{1}{9}$$$$1+2\times \frac{t^2-4t+3}{3}=\frac{1}{9}$$$$t^2-4t+3=-\frac{4}{3}$$$$t^2-4t+5=2-\frac{4}{3}=\frac{2}{3}$$$$\Rightarrow \sqrt{t^2-4t+5}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$$

Commented by jagoll last updated on 04/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

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