If 3 is a root of the equation {x^2} + kx - 2 | vedclass

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

Question

(C) Given that $3$ is a root of the quadratic equation ${x^2} + kx - 24 = 0.$Substituting $x = 3$ into the equation:${3^2} + k(3) - 24 = 0$$9 + 3k - 2

Vedclass · Accepted Answer

${x^2} - kx + 6 = 0$

Vedclass · Answer

(C) Given that $3$ is a root of the quadratic equation ${x^2} + kx - 24 = 0.$Substituting $x = 3$ into the equation:${3^2} + k(3) - 24 = 0$$9 + 3k - 24 = 0$$3k - 15 = 0$$3k = 15$$k = 5$Now,we check which equation is satisfied by $x = 3$ when $k = 5$:For option $(c)$: ${x^2} - kx + 6 = 0$Substituting $x = 3$ and $k = 5$:${3^2} - 5(3) + 6 = 9 - 15 + 6 = 0$Since the result is $0$,$3$ is a root of the equation ${x^2} - kx + 6 = 0$.

If $3$ is a root of the equation ${x^2} + kx - 24 = 0,$ then $3$ is also a root of which of the following equations?

Similar Questions

The positive value of $m$ for which the roots of the equation $12x^2 + mx + 5 = 0$ are in the ratio $3:2$ is

Find the value of $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(a-c)^{2}}{(a-b)(b-c)}$

The sum of all the real values of $x$ satisfying the equation $2^{(x - 1)(x^2 + 5x - 50)} = 1$ is

The least integral value $\alpha$ of $x$ such that $\frac{x - 5}{x^2 + 5x - 14} > 0$ satisfies:

If $\alpha, \beta$ are the roots of $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of $x^2 + qx + 1 = 0$,then $q^2 - p^2$ =

Vedclass Test Series

Exam Paper Generator

Online Exam Module