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

The number of the roots of the quadratic equation 8sec^2 θ − 6secθ + 1 = 0 is

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

Answered by mrW1 last updated on 20/Aug/17

8sec^2 θ − 6secθ + 1 = 0 (2sec θ − 1)(4sec θ − 1)= 0 sec θ=(1/2) ⇒ cos θ=2 ...(i) or sec θ=(1/4) ⇒ cos θ=4 ...(ii) (i) and (ii) have no solution ⇒number of roots is 0.

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

Commented by Tinkutara last updated on 20/Aug/17

Thank you very much Sir!

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

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