Find all $x$ numbers for $(x-1)(x-3) > 0$ I am learning calculus from Calculus by Michael Spivak and the exercise is to find all x numbers for $(x-1)(x-3) > 0$. Steps below are my procedure.


*

*$(x-1)(x-3) > 0$

*$x^2-4x+3 > 0$

*$x^2 > 4x-3$

*$x > +-\sqrt{4x-3}$


I was wondering if I can say that $x > +-\sqrt{4x-3}$ is the answer.

You haven't solved the inequality. You are looking for real valued solutions for $x$: values of $x$ that satisfy the inequality.
Keep the inequality in factored form.
$(x-1)(x-3) \gt 0$ is positive $(1)$ if both factors are positive, or $(2)$ if both factors are negative.


*

*when $(x-1)> 0$ and $(x - 3)>0 \implies x \gt 3.$

*when $(x - 1) < 0$ and $(x-3)\lt 0 \implies x \lt 1.$


So the inequality holds if $$x \in (-\infty, 1) \cup (3, \infty)$$
When $x \in [1, 3]$, the inequality fails.
