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.
Hello, fdrhs1984!
Did you try anything?
Are you waiting for Divine Intervention?
Find all real solutions of the equation: .$\displaystyle 10x^4 + 3x^2 \:=\:1$
We have: .$\displaystyle 10x^4 + 3x^2 - 1 \:=\:0$
Factor: .$\displaystyle (2x^2+1)(5x^2-1) \:=\:0$
And we have: .$\displaystyle \begin{Bmatrix}2x^2 + 1 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{1}{2} && \text{no real roots} \\ \\[-3mm]
5x^2 - 1 \:=\:0 & \Rightarrow & x^2 \:=\:\frac{1}{5} & \Rightarrow & x \:=\:\pm\frac{1}{\sqrt{5}} \end{Bmatrix}$