How many solutions are there to the equation $\sqrt{x^4+25}=x^2-5$?

Question

By squaring both sides we have:

$x^4+25=x^4-10x^2+25$

$10x^2=0$

$x=0$

Hence we have one solution to the equation. I'm not sure that what I've done is correct, in particular because it implies that $\sqrt{25}=-5$. Could you please tell me if I'm correct?

@Invisible oops! sorry deleted comment – Albus Dumbledore Apr 04 '21 at 10:09 — Albus Dumbledore, Apr 04 '21 at 10:09
$x=0$ here is called an extraneous solution. – ryang Apr 08 '21 at 07:21 — ryang, Apr 08 '21 at 07:21

score 1 · Accepted Answer · answered Apr 04 '21 at 09:27

1

Since $x=0$ doesn't satisfy the equation, it is not a solution. Hence there are no solutions.

Note that while taking the square root of a number, only absolute value is taken, that is, $\sqrt{25} \ne -5$.

answered Apr 04 '21 at 09:27

V.G

4,196

ok thank you very much – Michael Blane Apr 04 '21 at 09:29

score 1 · Answer 2 · answered Apr 04 '21 at 09:42

1

$\sqrt{x^4+25} =x^2-5$ implies $x^2 >5;$

Set $z:=x^2$ where $z >5;$

$\sqrt{z^2+25} > z >z - 5;$

No solutions to the equation.

answered Apr 04 '21 at 09:42

Peter Szilas

20,344
2
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How many solutions are there to the equation $\sqrt{x^4+25}=x^2-5$?

2 Answers2