The number of roots of the quadratic equation | vedclass

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(D) Given the quadratic equation $8\sec^2 	heta - 6\sec 	heta + 1 = 0$.Let $x = \sec 	heta$. The equation becomes $8x^2 - 6x + 1 = 0$.Factoring the

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$0$

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(D) Given the quadratic equation $8\sec^2 	heta - 6\sec 	heta + 1 = 0$.Let $x = \sec 	heta$. The equation becomes $8x^2 - 6x + 1 = 0$.Factoring the quadratic: $8x^2 - 4x - 2x + 1 = 0 \implies 4x(2x - 1) - 1(2x - 1) = 0$.This gives $(4x - 1)(2x - 1) = 0$.So,$x = \frac{1}{4}$ or $x = \frac{1}{2}$.Thus,$\sec 	heta = \frac{1}{4}$ or $\sec 	heta = \frac{1}{2}$.However,for any real $	heta$,the range of $\sec 	heta$ is $(-\infty, -1] \cup [1, \infty)$.Since both $\frac{1}{4}$ and $\frac{1}{2}$ lie in the interval $(-1, 1)$,there is no real value of $	heta$ that satisfies these equations.Therefore,the number of roots is $0$.

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

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