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Question Number 20001 by Tinkutara last updated on 20/Aug/17

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}$$$$\mathrm{equation}{8\sec }^2\theta -6\sec \theta +1=0\mathrm{is}$$

Answered by mrW1 last updated on 20/Aug/17

$${8\sec }^2\theta -6\sec \theta +1=0$$$$(2\sec \theta -1)(4\sec \theta -1)=0$$$$\sec \theta =\frac{1}{2}\Rightarrow \cos \theta =2...\left(\mathrm{i}\right)$$$$\mathrm{or}$$$$\sec \theta =\frac{1}{4}\Rightarrow \cos \theta =4...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{have}\mathrm{no}\mathrm{solution}$$$$$$$$\Rightarrow \mathrm{number}\mathrm{of}\mathrm{roots}\mathrm{is}0.$$

Commented by Tinkutara last updated on 20/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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