# How do you solve and write the following in interval notation: 2x^2-5x<=3?

Mar 20, 2017

$- \frac{1}{2} \le x \le 3 , \mathmr{and} , x \in \left[- \frac{1}{2} , 3\right] .$

#### Explanation:

$2 {x}^{2} - 5 x \le 3 \Rightarrow 2 {x}^{2} - 5 x - 3 \le 0.$

$\therefore \underline{2 {x}^{2} - 6 x} + \underline{x - 3} \le 0.$

$\therefore 2 x \left(x - 3\right) + 1 \left(x - 3\right) \le 0.$

$\therefore \left(x - 3\right) \left(2 x + 1\right) \le 0.$

This means that the Product of $2$ Real Nos. is $- v e .$

Therefore, $1$ No. must be $+ v e$ and the Other, $- v e$, or, vise-versa.

Accordingly, $2$ Cases arise :

Case (1) : $\left(x - 3\right) \ge 0 , \mathmr{and} , \left(2 x + 1\right) \le 0.$

:. x ge 3, &, x le -1/2, which is Absurd.

Hence, this Case is Imossible.

Case (2) : $\left(x - 3\right) \le 0 , \mathmr{and} , \left(2 x + 1\right) \ge 0.$

In this Case, $x \le 3 , \mathmr{and} , x \ge - \frac{1}{2} , \text{ i.e., to say, } - \frac{1}{2} \le x \le 3.$

Thus, Desired Interval Notation, is,

$- \frac{1}{2} \le x \le 3 , \mathmr{and} , x \in \left[- \frac{1}{2} , 3\right] .$

Enjoy Maths.!