The number of roots of the quadratic equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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

Similar Questions

The principal solutions of $\sqrt{3} \sec x + 2 = 0$ are

If $2\sin^2 \theta = 3\cos \theta$,where $0 \le \theta \le 2\pi$,then $\theta = $

The values of $x$ satisfying the equation $3 \operatorname{cosec} x = 4 \sin x$ are

Let $S = \{ \theta \in [ - 2\pi , 2\pi ] : 2\cos^2 \theta + 3\sin \theta = 0 \}$. Then the sum of the elements of $S$ is

If $\sqrt{3}(\cos ^{2} x)=(\sqrt{3}-1) \cos x+1,$ the number of solutions of the given equation when $x \in [0, \frac{\pi}{2}]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module