35. $x^4-5x^2+4<0$

Key takeaway: This quartic is a quadratic in $x^2$. Rewriting it that way makes the inequality much easier to solve.

Let

$t=x^2$

Then the inequality becomes

$t^2-5t+4<0$

Factor:

$t^2-5t+4=(t-1)(t-4)$

So we need

$(t-1)(t-4)<0$

This product is negative between the roots:

$1<t<4$

Now substitute back $t=x^2$:

$1<x^2<4$

That means

$1<|x|<2$

So the solution is

$(-2,-1)\cup(1,2)$

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Pitfalls the pros know 👇 Do not stop at $1<x^2<4$ and write $1<x<4$. The variable is $x^2$, so you must convert the result back to $x$ by considering both positive and negative values.

What if the problem changes? If the inequality were $x^4-5x^2+4\le 0$, then the endpoints would also be included:

$x\in[-2,-1]\cup[1,2]$

If it were $>0$, the solution would be the outside regions:

$x\in(-\infty,-2)\cup(-1,1)\cup(2,\infty)$

Tags: quartic inequality, substitution, factorization