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

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(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

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

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(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

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