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

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Answer 1 · 2017-03-20T13:13:36

# -1/2 le x le 3, or, x in [-1/2,3].#

#2x^2-5x le 3 rArr 2x^2-5x-3 le 0.#

#:. ul(2x^2-6x)+ul(x-3) le 0.#

# :. 2x(x-3)+1(x-3) le 0.#

#:. (x-3)(2x+1) le 0.#

This means that the Product of #2# Real Nos. is #-ve.#

Therefore, #1# No. must be #+ve# and the Other, #-ve#, or, vise-versa.

Accordingly, #2# Cases arise :

Case (1) : #(x-3) ge 0, and, (2x+1) le 0.#

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

Hence, this Case is Imossible.

Case (2) : #(x-3) le 0, and, (2x+1) ge 0.#

In this Case, #x le 3, and, x ge -1/2," i.e., to say, "-1/2 le x le 3.#

Thus, Desired Interval Notation, is,

# -1/2 le x le 3, or, x in [-1/2,3].#

Enjoy Maths.!