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.
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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)
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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
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