If alpha neq 0 and 0 are the roots of the equ | vedclass

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

Question

(C) Given the quadratic equation $x^2 - 5kx + (6k^2 - 2k) = 0$. Since $0$ is a root of the equation,it must satisfy the equation: $(0)^2 - 5k(0) + (6k

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(C) Given the quadratic equation $x^2 - 5kx + (6k^2 - 2k) = 0$. Since $0$ is a root of the equation,it must satisfy the equation: $(0)^2 - 5k(0) + (6k^2 - 2k) = 0$ $6k^2 - 2k = 0$ $2k(3k - 1) = 0$ This gives $k = 0$ or $k = \frac{1}{3}$. If $k = 0$,the equation becomes $x^2 = 0$,which has roots $0, 0$. Since $\alpha 
eq 0$,$k$ cannot be $0$. If $k = \frac{1}{3}$,the equation becomes $x^2 - 5(\frac{1}{3})x + (6(\frac{1}{3})^2 - 2(\frac{1}{3})) = 0$ $x^2 - \frac{5}{3}x + (6(\frac{1}{9}) - \frac{2}{3}) = 0$ $x^2 - \frac{5}{3}x + (\frac{2}{3} - \frac{2}{3}) = 0$ $x^2 - \frac{5}{3}x = 0$ $x(x - \frac{5}{3}) = 0$ The roots are $0$ and $\frac{5}{3}$. Since $\alpha$ is the non-zero root,$\alpha = \frac{5}{3}$.

If $\alpha \neq 0$ and $0$ are the roots of the equation $x^2 - 5kx + (6k^2 - 2k) = 0$,then $\alpha = $

Similar Questions

If $x^2 + x + 1$ is a factor of $ax^3 + bx^2 + cx + d$,then what is the real root of $ax^3 + bx^2 + cx + d = 0$?

Sum of all the roots of the equation $||2x-3|-4|=2$ is

The polynomial equation of degree $4$ having real coefficients with three of its roots as $2 \pm \sqrt{3}$ and $1+2i$ is:

Let $p, q, r \in \mathbb{R}$ and $r > p > 0$. If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta$,then what is the value of $|\alpha| + |\beta|$?

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation $6x^2 - 11x + \alpha = 0$ are rational numbers is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module