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.
-
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)
-
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
Related Maths A Level answers
All answers ▸How do you find the coordinates of stationary points on a graph?
How can we calculate the derivative of function f(x)= (x+2)/(x-1)?
Find the equation of the tangent to the curve y=x^2+5x+2 at the point where x=5
In a triangle ABC, side AB=10 cm, side AC=5cm and the angle BAC=θ, measured in degrees. The area of triangle ABC is 15cm(sq). Find 2 possible values for cosθ and the exact length of BC, given that it is the longest side of the triangle.