Solve-x-2-3x-x-2-2-0- Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 21355 by Tinkutara last updated on 21/Sep/17 Solve:∣x2+3x∣+x2−2⩾0 Answered by mrW1 last updated on 22/Sep/17 ∣x2+3x∣⩾2−x21.x2+3x⩾0{x2+3x⩾0⇒x(x+3)⩾0x2+3x⩾2−x2⇒2x2+3x−2⩾0⇒(2x−1)(x+2)⩾0⇒{x⩾0orx⩽−3x⩾12orx⩽−2⇒x⩾12orx⩽−32.x2+3x⩽0{x2+3x⩽0⇒x(x+3)⩽0⇒−3⩽x⩽0−(x2+3x)⩾2−x2⇒−3x⩾2⇒x⩽−23⇒−3⩽x⩽−23⇒x∈(−∞,−23]or[12,+∞) Commented by Tinkutara last updated on 22/Sep/17 ThankyouverymuchSir! Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Solve-2x-5-x-1-gt-8-Next Next post: Question-152431 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![∣x^2 +3x∣≥2−x^2 1. x^2 +3x≥0 { ((x^2 +3x≥0 ⇒x(x+3)≥0)),((x^2 +3x≥2−x^2 ⇒2x^2 +3x−2≥0 ⇒(2x−1)(x+2)≥0)) :} ⇒ { ((x≥0 or x≤−3)),((x≥(1/2) or x≤−2)) :} ⇒x≥(1/2) or x≤−3 2. x^2 +3x≤0 { ((x^2 +3x≤0 ⇒x(x+3)≤0 ⇒−3≤x≤0)),((−(x^2 +3x)≥2−x^2 ⇒−3x≥2 ⇒x≤−(2/3))) :} ⇒−3≤x≤−(2/3) ⇒x∈(−∞,−(2/3)] or [(1/2),+∞)](https://www.tinkutara.com/question/Q21381.png)
