Solving $\sqrt{|1-2x|}\geq1+x$

Question

Where do I start when solving $\sqrt{|1-2x|}\geq1+x$? Is squaring both sides allowed?

Answer 1

for $$x\le -1$$ the inequality is true. For $$x\geq -1$$ we can square it $$|1-2x|\geq (x+1)^2$$ squaring again and factorizing we get $$-x \left( x+4 \right) \left( {x}^{2}+2 \right) \geq 0$$ we only need to discuss $$-x(x+4)\geq 0$$ we we can conclude that $$-4\le x\le 0$$ with $$x\geq -1$$ we get $$-1\le x\le 0$$ thus we get the solution set $$x\le 0$$

Answer 2

It is allowed but you need to set conditions on the expression under the square root, i.e. for $1-2x$. This should be done because you only can take square roots of numbers $\gt 0$.

Answer 3

Hint: if $1 + x\le 0$, the inequality is obviously true, And if $1 + x > 0$ squaring both sides is allowed because...

Answer 4

If $1-2x\ge0\iff x\le\dfrac12\ \ \ \ (1)$ we have $$1+x\le\sqrt{1-2x}$$

As $\sqrt{1-2x}\ge0,$ the relationship holds implicitly if $1+x<0$

Now square both sides for $x+1\ge0$ and ensure $(1)$ is satisfied.

What if $1-2x<0\iff x>\dfrac12?$