Question

6x^4 - 18x^2 +6 = 0
Solve for 0.

Answer 1

Simplifying
6x4 + -18x2 + 6 = 0

Reorder the terms:
6 + -18x2 + 6x4 = 0

Solving
6 + -18x2 + 6x4 = 0

Solving for variable 'x'.

Factor out the Greatest Common Factor (GCF), '6'.
6(1 + -3x2 + x4) = 0

Ignore the factor 6

Answer 2

How do I solve x^4-3x^2+1?

Answer 3

@tanya123

Answer 4

Let \(u=x^2\). Rewrite the equation in terms of \(u\).

Answer 5

a^2 - 3a+1

Answer 6

or u^2 - 3u +1

Answer 7

Use the quadratic equation.

Answer 8

ooo right

Answer 9

Nothing happen to the u's being back to x^2?
I checked the solutions for 6x^4-18x^2+6=0 and I get 4 solutions.

Answer 10

@thomas5267

Answer 11

but when I solve for u I get only 2 solutions

Answer 12

Note that if \(x^2=n\), then \(x=\sqrt{n}\) and \(x=-\sqrt{n}\).

Answer 13

I got confused by that

Answer 14

Consider this case. If \(x^2=9\), clearly \(x=3\) is a solution. However, \(x=-3\) is also a solution as \((-3)^2=9\).

Answer 15

oh yes

Answer 16

but in the case after getting the solutions of u after using the quadratic formula. What do I do to get the other 2?

Answer 17

I got x= 1/2(3-sqrt(5)) and x=1/2(3+sqrt(5))

Answer 18

after solving for u

Answer 19

but the real solutions that I need from 6x^4 -18x^2 +6 = 0 are 4

Answer 20

\[
6(x^4-3x^2+1)=0\\
u=x^2\\
\begin{align*}
u^2-3u+1&=0\\
u&=\frac{3\pm\sqrt{(-3)^2-4}}{2}\\
u&=\frac{3\pm\sqrt{5}}{2}\\
u&=\frac{3+\sqrt{5}}{2} \text{ or } u=\frac{3-\sqrt{5}}{2}
\end{align*}
\]

Answer 21

ooo

Answer 22

the u's become x^2 and then I get 2 more solutions. But how do I simplify the square root over the solutions of u?

Answer 23

\[
\begin{align*}
u&=\frac{3+\sqrt{5}}{2}&&\\
x^2&=\frac{3+\sqrt{5}}{2}&\text{ or } x^2&=\frac{\sqrt{5}-3}{2}\\
x&=\sqrt{\frac{3+\sqrt{5}}{2}}&\text{ or } x&=\sqrt{\frac{\sqrt{5}-3}{2}}\\
\end{align*}
\]
The problem now is how to simplify the square root in square root.

Answer 24

yes. I got that. I'm trying to simplify that now but I don't know how. I haven't been taught how to do that

Answer 25

\[x=\frac{ 1+\sqrt{5} }{ 2 }\]

Answer 26

How can I simplify it to that?

Answer 27

I'm aware there are 3 more

Answer 28

I found this document on the internet. This document is about how to solve nested square roots.

http://www.cybertester.com/data/denest.pdf

Answer 29

@ganeshie8 How to solve nested square roots? I need a coffee break now lol.

Answer 30

Theorem (4.12) in that document might be helpful.

Answer 31

Formula given in the link:
\[
\sqrt{A\pm B}=\sqrt{\frac{1}{2}A+\frac{1}{2}\sqrt{A^2-B^2}}\pm\sqrt{\frac{1}{2}A-\frac{1}{2}\sqrt{A^2-B^2}} \text{, given } A,B\in\mathbb{R} \\\text{ and }A>B>0
\]

\[
\begin{align*}
x=\sqrt{\frac{3}{2}+\frac{\sqrt{5}}{2}}&=\sqrt{\frac{3}{4}+\frac{1}{2}\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}^2\right)}}+\sqrt{\frac{3}{4}-\frac{1}{2}\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}^2\right)}}\\
&=\sqrt{\frac{3}{4}+\frac{1}{2}\sqrt{\frac{9}{4}-\frac{5}{4}}}+\sqrt{\frac{3}{4}-\frac{1}{2}\sqrt{\frac{9}{4}-\frac{5}{4}}}\\
&=\sqrt{\frac{3}{4}+\frac{1}{2}}+\sqrt{\frac{3}{4}-\frac{1}{2}}\\
&=\sqrt{\frac{5}{4}}+\sqrt{\frac{1}{4}}\\
&=\frac{\sqrt{5}}{2}+\frac{1}{2}\\
&=\frac{1+\sqrt{5}}{2}
\end{align*}
\]

Answer 32

<3

Answer 33

In case you can't read the overflowed part.
\[
\begin{align*}
x&=\sqrt{\frac{3}{2}+\frac{\sqrt{5}}{2}}\\
&=\sqrt{\frac{3}{4}+\frac{1}{2}\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}^2\right)}}+\sqrt{\frac{3}{4}-\frac{1}{2}\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}^2\right)}}\\
&=\sqrt{\frac{3}{4}+\frac{1}{2}\sqrt{\frac{9}{4}-\frac{5}{4}}}+\sqrt{\frac{3}{4}-\frac{1}{2}\sqrt{\frac{9}{4}-\frac{5}{4}}}\\
&=\sqrt{\frac{3}{4}+\frac{1}{2}}+\sqrt{\frac{3}{4}-\frac{1}{2}}\\
&=\sqrt{\frac{5}{4}}+\sqrt{\frac{1}{4}}\\
&=\frac{\sqrt{5}}{2}+\frac{1}{2}\\
&=\frac{1+\sqrt{5}}{2}
\end{align*}
\]

Answer 34

I am probably overcomplicating things as usual, @ganeshie8 . It is absolutely common for me to do so. Any other solutions?

Answer 35

You now have to do this to all four solutions... Good luck... :( I have to take a coffee break.

Answer 36

Yeah, I got it. I am falling sleep but still work to do haha

Answer 37

Thanks you very much. After this I will have an even more problematic one. Idk if anyone would be able to help me out on that one.

Answer 38

@thomas5267 I have a question regarding this again. When I solve for sqrt(3-sqrt(5)/2) wht changes on the equation?

Answer 39

the above one is 3/2 + sqrt5/2 but now i have 3/2 - sqrt5/2