x2−2x−15≥0 x2−5x+3x−15≥0 x(x−5)+3(x−5)≥0 (x−5)(x+3)≥0 5 and -3 are critical points hence, x∈(−∞,−3]∪[5,∞)

Resolved: "(x2)−2x−15→0"

Secondary
Algebra
Algebra I
Exponents and radicals

Emily-Jane Bray

Answered question

2021-02-15

(x^{2}) - 2 x - 15 \to 0

Answer & Explanation

Jayden-James Duffy

Skilled2021-02-16Added 91 answers

x^{2} - 2 x - 15 \geq 0

x^{2} - 5 x + 3 x - 15 \geq 0

x (x - 5) + 3 (x - 5) \geq 0

(x - 5) (x + 3) \geq 0

5 and -3 are critical points hence,

x \in (- \infty, - 3] \cup [5, \infty)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?

Product

High School Questions
College Questions
Math Solver
Top Questions
Top Questions 2

Company

About us
Term of Service
Privacy
Payment Policy
Honor

Connect with us

Youtube
Facebook

Get Plainmath App

Google Play

Louki Akrita, 23, Bellapais Court, Flat/Office 46, 1100, Nicosia, Cyprus

Cyprus reg.number: ΗΕ 419361

E-mail us: support@plainmath.net

Our Service is useful for:

Plainmath is a platform aimed to help users to understand how to solve math problems by providing accumulated knowledge on different topics and accessible examples. Plainmath.net is owned and operated by RADIOPLUS EXPERTS LTD.