If both roots of the quadratic equation x^2 + | vedclass

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

Question

(B) For the roots of $x^2 + bx + c = 0$ to be positive and distinct,we need: $1$. Discriminant $D > 0$ $2$. Sum of roots $-b > 0$ $3$. Product of root

Vedclass · Accepted Answer

$\left( \frac{13\pi}{12}, \frac{17\pi}{12} \right)$

Vedclass · Answer

(B) For the roots of $x^2 + bx + c = 0$ to be positive and distinct,we need: $1$. Discriminant $D > 0$ $2$. Sum of roots $-b > 0$ $3$. Product of roots $c > 0$ Given $x^2 + (\sin 	heta + \cos 	heta)x + \frac{3}{8} = 0$: Product of roots = $\frac{3}{8} > 0$ (Always true). Sum of roots = $-(\sin 	heta + \cos 	heta) > 0 \implies \sin 	heta + \cos 	heta Discriminant $D = (\sin 	heta + \cos 	heta)^2 - 4(\frac{3}{8}) > 0$ $(\sin 	heta + \cos 	heta)^2 > \frac{3}{2}$ $1 + \sin 2	heta > \frac{3}{2} \implies \sin 2	heta > \frac{1}{2}$. Solving $\sin 2	heta > \frac{1}{2}$ for $2	heta \in [0, 4\pi]$: $\frac{\pi}{6} $\frac{\pi}{12} Check condition $\sin 	heta + \cos 	heta For $	heta \in (\frac{\pi}{12}, \frac{5\pi}{12})$,$\sin 	heta + \cos 	heta > 0$ (Both positive). For $	heta \in (\frac{13\pi}{12}, \frac{17\pi}{12})$,$\sin 	heta + \cos 	heta Thus,the solution is $	heta \in (\frac{13\pi}{12}, \frac{17\pi}{12})$.

If both roots of the quadratic equation $x^2 + (\sin \theta + \cos \theta)x + \frac{3}{8} = 0$ are positive and distinct,then the complete set of values of $\theta$ in $[0, 2\pi]$ is:

Similar Questions

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$

Two roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$ are positive and equal. If the product of the other two real roots is $1$,then:

If the figure shows the graph of $y = ax^2 + bx + c$,then . . . . . .

The sum of two roots of the equation $x^4-x^3-16x^2+4x+48=0$ is zero. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\alpha^4+\beta^4+\gamma^4+\delta^4=$

Let $\alpha, \beta \in \mathbb{N}$ be roots of the equation $x^2-70x+\lambda=0$,where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbb{N}$. If $\lambda$ assumes the minimum possible value,then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module