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

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given inequality: $2x^2 + 3x - 9 \le 0$Factorizing the quadratic expression: $2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3) \le 0$$(2x - 3)(x + 3) \le 0$The roots of the equation $2x^2 + 3x - 9 = 0$ are $x = 3/2$ and $x = -3$.For a quadratic expression $ax^2 + bx + c \le 0$ where $a > 0$,the solution lies between the roots,i.e.,$	ext{root}_1 \le x \le 	ext{root}_2$.Thus,the solution is $-3 \le x \le 3/2$.

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

Similar Questions

If $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1$,then find the value of $z+\frac{1}{x}$.

What is the sum of the squares of the roots of the quadratic equation $x^2 - 3x + 1 = 0$?

The number of solutions of $\log_4(x - 1) = \log_2(x - 3)$ is:

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is:

The value of $k$ for which the quadratic equation $(k + 11)x^2 - (k + 3)x + 1 = 0$ has real and equal roots is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module