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

The sum of all the roots of the equation $|x^2-8x+15|-2x+7=0$ is:

The number of ordered pairs $(a, b)$ of integers such that $a-b$ is a root of $x^2+ax+b=0$ is

If $\left|\frac{x^2+kx+1}{x^2+x+1}\right| < 3$ for all real $x$,then $k$ is in the interval

If $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of the equation $x^2 + 2bx + b = 0$,and $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of the equation $x^2 + 4x + 2 = 0$,then find the value of $b$.

Let the transformed equation of $2x^4-8x^3+3x^2-1=0$ such that the term containing the cubic power of $x$ is absent be $2x^4+bx^2+cx+d=0$. Then $b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module