$$2 \sqrt{x-1} < x $$

Please help me in solving this elementary inequality problem.

I can figure out that since the LHS $\ge$ $0$, $x > 0$ (1) and since the term within the root has to be greater than zero, $x \ge 1$. However, 2 is clearly not acceptable, but there is no way that I can figure how to proceed from here.

Any help pointing to a useful link would also be greatly appreciated. I am having difficulties with solving rational inequalities, specially ones that involve a quadratic expression with roots/modulus, and I don't have a text book with me as of now. Any online link helping me out wouuld be just great! Thanks !!


$x\ge 1$ is necessary and both terms must be positive so that the inequality is equivalent to this one : $4(x-1)<x^2$ that is $(x-2)^2>0$

I'll let you conclude with a picture :


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.