# Find All Real Solutions

• September 13th 2010, 09:14 AM
fdrhs1984
Find All Real Solutions
Find all real solutions of the equation exactly.

10x^4 + 3x^2 = 1
• September 13th 2010, 09:16 AM
undefined
Quote:

Originally Posted by fdrhs1984
Find all real solutions of the equation exactly.

10x^4 + 3x^2 = 1

What have you tried?

Rule # 11, show some effort.

http://www.mathhelpforum.com/math-he...ng-151421.html

That said, if you substitute y = x^2 you will get a quadratic equation.
• September 13th 2010, 09:33 AM
Soroban
Hello, fdrhs1984!

Did you try anything?
Are you waiting for Divine Intervention?

Quote:

Find all real solutions of the equation: . $10x^4 + 3x^2 \:=\:1$

We have: . $10x^4 + 3x^2 - 1 \:=\:0$

Factor: . $(2x^2+1)(5x^2-1) \:=\:0$

And we have: . $\begin{Bmatrix}2x^2 + 1 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{1}{2} && \text{no real roots} \\ \\[-3mm]
5x^2 - 1 \:=\:0 & \Rightarrow & x^2 \:=\:\frac{1}{5} & \Rightarrow & x \:=\:\pm\frac{1}{\sqrt{5}} \end{Bmatrix}$

• September 14th 2010, 06:27 AM
fdrhs1984
Quote:

Originally Posted by Soroban
Hello, fdrhs1984!

Did you try anything?
Are you waiting for Divine Intervention?

We have: . $10x^4 + 3x^2 - 1 \:=\:0$

Factor: . $(2x^2+1)(5x^2-1) \:=\:0$

And we have: . $\begin{Bmatrix}2x^2 + 1 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{1}{2} && \text{no real roots} \\ \\[-3mm]
5x^2 - 1 \:=\:0 & \Rightarrow & x^2 \:=\:\frac{1}{5} & \Rightarrow & x \:=\:\pm\frac{1}{\sqrt{5}} \end{Bmatrix}$

I don't understand how you got (2x^2 + 1)(5x^2 - 1).
• September 14th 2010, 07:02 AM
CaptainBlack
Quote:

Originally Posted by fdrhs1984
I don't understand how you got (2x^2 + 1)(5x^2 - 1).

Put $y=x^2$ then you have a quadratic in $y$ to solve, from the solutions you can then find the $x$'s that solve the original equation.

CB