Solve inequality: sqrt(x^2) + x < 1

There are two possible solutions, substitution and using absolute value. Here, we will use absolute value method.By definition of square root we get: sqrt(x^2) = |x|. By definition of absolute value we get:|x| = x, if x >= 0 and |x| = -x, if x<0 (Definition 1). Rewrite inequality as: |x| + x < 1 Thus, we need to observe two possible options, where x is non-negative and negative.

  1. x >= 0, thus x + x < 1 (By Definition 1) x >= 0, 2x < 1 x>= 0 , x < 1/2 The range for this system is [0; 1/2)

  2. x < 0, thus -x + x < 1 0 < 1 (always true) The range for this system is (-infinity, 0)

We combine these two intervals and get final answer (-infinity, 1/2)

RF
Answered by Robert F. Maths tutor

6938 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The first term of an infinite geometric series is 48. The ratio of the series is 0.6. (a) Find the third term of the series. (b) Find the sum to infinity. (c) The nth term of the series is u_n. Find the value of the sum from n=4 to infinity of u_n.


How do you intergrate a function?


What is the gradient of the function f(x) = 2x^2 + 3x - 7 at the point where x = -2?


Q4 on 2017 Edexcel C4 paper, concerns differentiation of multiple variables.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning