Let alpha, beta be the roots of the equation | vedclass

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

Question

(C) Given the equation $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) = 1$. This implies $\frac{x^2}{\beta^2} = |\alpha|$,so $x^2 = \beta^2 |\alp

Vedclass · Accepted Answer

$< \beta$

Vedclass · Answer

(C) Given the equation $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) = 1$. This implies $\frac{x^2}{\beta^2} = |\alpha|$,so $x^2 = \beta^2 |\alpha|$. Since $\alpha, \beta$ are roots of $x^2 - |a|x - |b| = 0$,the product of roots $\alpha \beta = -|b| \le 0$. Given $|\alpha|  0$,we analyze the signs. Since $|a| Solving for $x$,we get $x = \pm |\beta| \sqrt{|\alpha|}$. The positive root is $|\beta| \sqrt{|\alpha|}$. Since $|\alpha| Since $\beta > 0$ and $\sqrt{|\alpha|} < 1$ (as $|\alpha| < |\beta|$ and the product of roots is negative),the positive root is less than $\beta$.

Let $\alpha, \beta$ be the roots of the equation $x^2 - |a|x - |b| = 0$ such that $|\alpha| < |\beta|$. If $|a| < \beta - 1$,then the positive root of $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) - 1 = 0$ is

Similar Questions

Evaluate $\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$ for $x \ge 1$.

If $m$ is chosen in the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$ such that the sum of its roots is greatest,then the absolute difference of the cubes of its roots is

If the quadratic equation $x^2 + (2 - \tan \theta)x - (1 + \tan \theta) = 0$ has $2$ integral roots,then the sum of all possible values of $\theta$ in the interval $(0, 2\pi)$ is $k\pi$,then $k$ equals:

Solve $|x - 2| + |x - 1| = x - 3$

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 $\alpha^4+\beta^4+\gamma^4+\delta^4=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module