Solve the inequality $x^4-3x^2+5\ge0$

Question

Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$

$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x^2+5\ge0.$$Let $x^2=y,y\ge0.$ Now we have $$y^2-3y+5\ge0; D=9-4\times5<0,a=1>0$$ so the function $f(y)=y^2-3y+5>0$ for all $y$. I don't know what to do next. The solution of the inequality is indeed $x\in(-\infty;+\infty),$ but I have the restriction $y\ge0?$

To solve $x^4-3x^2+5\ge0$, just let $X=x^2$. Then solve $X^2-3X+5\ge0$, and finally come back to $x$. — TheSilverDoe, Oct 21 '20 at 15:26
Note that $0\le (x^2-\frac32)^2=x^4-3x^2+\frac 94< x^4-3x^2+5$ — Hagen von Eitzen, Oct 21 '20 at 15:29
Hint : multiply your equation by :$\sqrt{x^4-3x^2+12}-\sqrt{x^4-3x^2+5}$. — Kelenner, Oct 21 '20 at 15:33
@Kelenner, I don't really see how that helps; in effect it only replaces the $7$ with a $1$. — Barry Cipra, Oct 21 '20 at 15:38
@Barry Cipra Yes, but you have then the difference $\sqrt{x^4-3x^2+12}-\sqrt{x^4-3x^2+5}$ equal to $1$. Add with $\sqrt{x^4-3x^2+12}+\sqrt{x^4-3x^2+5}=7$... — Kelenner, Oct 21 '20 at 15:42

cosmo5 · Answer 1 · 2020-10-21T15:51:25.877

5

Hint :

First solve for $\sqrt{z} + \sqrt{z+7} = 7 $

Next substitute $z=x^4-3x^2+5$ and solve for $x$.

Actually easy to see $\sqrt{9} + \sqrt{9+7} = 7 $.

$9$ is only solution as monotonic function has unique root (when it does).

So only need to solve for $x^4-3x^2+5=9$.

edited Oct 21 '20 at 15:51

answered Oct 21 '20 at 15:26

cosmo5

10,629

peter.petrov · Accepted Answer · 2020-10-21T15:34:53.137

Denote:

$x^4-3x^2+5 = a - 7/2$

Then:

$x^4-3x^2+12 = a + 7/2$

Now you want to solve this one:

$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$

This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.

At the end do a direct check to see if the values you found for $a$ give rise to valid roots for $x$. For example, if you get a solution for $a$ smaller than $7/2$, that obviously does not work i.e. does not give you any solutions for $x$.

Solve the inequality $x^4-3x^2+5\ge0$

2 Answers2