The set {x in R: 4+11x-3x^20} is the interva | vedclass

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

Question

(A) Given inequality: $4+11x-3x^2 > 0$ Multiply by $-1$ on both sides,the inequality sign reverses: $3x^2 - 11x - 4 < 0$ Factorize the quadratic expre

Vedclass · Accepted Answer

$\left(-\frac{1}{3}, 4\right)$

Vedclass · Answer

(A) Given inequality: $4+11x-3x^2 > 0$ Multiply by $-1$ on both sides,the inequality sign reverses: $3x^2 - 11x - 4 Factorize the quadratic expression: $3x^2 - 12x + x - 4 $3x(x - 4) + 1(x - 4) $(3x + 1)(x - 4) To find the interval where the product is negative,we find the roots: $x = -\frac{1}{3}$ and $x = 4$. Testing the intervals: For $x For $-\frac{1}{3} For $x > 4$,the expression is positive. Therefore,the solution set is $x \in \left(-\frac{1}{3}, 4\right)$.

The set $\{x \in R: 4+11x-3x^2>0\}$ is the interval

Similar Questions

If $x$ is a real number,for what values of $x$ does $3x^2 + 14x + 11 > 0$ hold?

The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by

Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$, then

Assertion $(A)$: $3x^2 - 16x + 4 > -16$ is satisfied for some values of real $x$ in $(0, \frac{10}{3})$.
Reason $(R)$: $ax^2 + bx + c$ and $a$ will have the same sign for some values of $x \in \mathbb{R}$ when $b^2 - 4ac > 0$.
The correct option among the following is

If $x^2+2px-2p+8>0$ for all real values of $x$,then the set of all possible values of $p$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module