Question

下列哪组 $x$ 值满足不等式 $2x^2+x\lt6$ ？

Which of the following sets of $x$ -values satisfy the inequality $2x^2+x\lt6?$

$-2\lt x\lt\dfrac32$

$x\gt\dfrac32$ 或 $x\lt-2$

$x\gt\dfrac32$ or $x\lt-2$

$x\lt\dfrac32$

$\dfrac32\lt x\lt2$

$x\lt-2$

将所有项移到一边，并分解二次式

Move all terms to one side and factor the quadratic

两个一次因式的乘积在两根之间为负

A product of two linear factors is negative between its roots

该不等式为 $2x^2+x-6\lt0$ 。分解得 $(2x-3)(x+2)\lt0$ 。乘积在两根之间为负，所以 $-2\lt x\lt\frac{3}{2}$ 。

所以正确答案是 A。

The inequality is $2x^2+x-6\lt0.$ Factoring gives $(2x-3)(x+2)\lt0.$ The product is negative between its roots, so $-2\lt x\lt\frac{3}{2}.$

Thus, the correct answer is A.