The solution of the equation 2x^2 + 3x - 9 le | vedclass

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

Question

(B) Given the quadratic inequality: $2x^2 + 3x - 9 \le 0$Step $1$: Factorize the quadratic expression.$2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3)

Vedclass · Accepted Answer

$-3 \le x \le 3/2$

Vedclass · Answer

(B) Given the quadratic inequality: $2x^2 + 3x - 9 \le 0$Step $1$: Factorize the quadratic expression.$2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3) \le 0$$(2x - 3)(x + 3) \le 0$Step $2$: Find the critical points by setting the factors to zero.$2x - 3 = 0 \implies x = 3/2$$x + 3 = 0 \implies x = -3$Step $3$: Determine the interval where the product is less than or equal to zero.For a quadratic $(x - \alpha)(x - \beta) \le 0$ where $\alpha Here,$-3 \le x \le 3/2$.Thus,the correct option is $B$.

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

Similar Questions

The solution set of the inequation $\sqrt{x^2+6x+5} > (8-x)$ is

If $a$ and $b$ are the roots of the equation $x^2-7x-1=0$,then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to $........$.

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

Let $f(x) = x^2 + 4x + 1$. Then

The solution set of the inequation $3^x+3^{1-x}-4 < 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module