How to solve a twin root equation?

Question

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I thought to factorize a sqrt(x), but I can't find out anything. I thought to multiply both sides with themselves four times, but I'm not sure that works.

This isn't a differential equation. – Shahar Feb 25 '15 at 18:10 — Shahar, Feb 25 '15 at 18:10
Hint: it is $, X^2! + X = 12\ $ for $\ X = x^{1/4}\ \ $ – Bill Dubuque Feb 25 '15 at 18:22 — Bill Dubuque, Feb 25 '15 at 18:22

score 2 · Answer 1 · answered Feb 25 '15 at 18:13

2

HINT:

$x^{1/2}+x^{1/4}=x^{1/4}(x^{1/4}+1)$

now let

$y=x^{1/4}$

amd solve $y(y+1)=12$

answered Feb 25 '15 at 18:13

Shobhit

6,902

You beat me to it. – John Feb 25 '15 at 18:14
@John hey we are on the same team ;p – Shobhit Feb 25 '15 at 18:17

score 2 · Answer 2 · answered Feb 25 '15 at 18:15

2

You can take out an $x^\frac14$:

$$x^\frac14\left(x^\frac14+1\right)=12$$ And let $u=x^\frac14$, and thus you have:

$$u(u+1)=12$$

Obviously, $u=3$ (but you can also use quadratic formula to find this), so:

$$3=x^\frac14\\\therefore x=81$$

answered Feb 25 '15 at 18:15

Shahar

3,302

How to solve a twin root equation?

2 Answers2