Solve x⁴-5x²+4=0: Finding All Solutions to a Fourth-Degree Equation

Let's solve the given equation:

$x^4-5x^2+4=0$

We identify that this is a bi-quadratic equation that can be easily solved using substitution of a new variable,

That is, let's notice that:

$(x^2)^2=x^4$

Therefore, we can write the given equation in the following form:$\textcolor{blue}{ x^4}-5x^2+4=0 \\ \downarrow\\ \textcolor{blue}{ (x^2)^2}-5x^2+4=0$

Now let's define a new variable, $t$, such that:

$t=x^2$

If we substitute this new variable, $t$, in the given equation instead of $x^2$ we'll obtain an equation that depends only on $t$:

$(\textcolor{red}{x^2})^2-5\textcolor{red}{x^2}+4=0 \\ \downarrow\downarrow \boxed{\textcolor{red}{(x^2=t)}}\\ t^2-5t+4=0$

Proceed to solve the new equation that we obtained for the variable $t$. After we determine the values of variable t for which the equation holds, we'll go back and substitute each of them into the definition of t that we mentioned before in order to determine the value of x,

We identify that the equation that we obtained in the last step for t is a quadratic equation that can be solved using quick trinomial factoring:

$t^2-5t+4=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=4\\ \underline{?}+\underline{?}=-5\end{cases}\\ \downarrow\\ (t-4)(t-1)=0$

Therefore we'll obtain two simpler equations from which we'll extract the solution for t:

$(t-4)(t-1)=0 \\ \downarrow\\ t-4=0\rightarrow\boxed{t=4}\\ t-1=0\rightarrow\boxed{t=1}\\ \boxed{t=1,4}$

Now let's go back to the definition of t that was mentioned before, let's recall it:

$t=x^2$ Notice that given the power of x is even, the variable t can get only non-negative values (meaning positive or zero),

Therefore the two values that we obtained for t from solving the quadratic equation are indeed valid,

We'll continue to substitute each of the two values we that we obtained for t in the definition of t mentioned before to solve the equation and then proceed to extract the corresponding value of x by solving the resulting equation using square root on both sides:

$\boxed{x^2=t}\\ \downarrow\\ t=1\textcolor{blue}{\rightarrow} x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm1}\\ t=4\textcolor{blue}{\rightarrow} x^2=4\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm2}\\ \downarrow\\ \boxed{x=\pm1,\pm2}$

Let's summarize the steps of solving the equation:

$x^4-5x^2+4=0 \\ \downarrow\\ (\textcolor{red}{x^2})^2-5\textcolor{red}{x^2}+4=0 \\ \downarrow\downarrow \boxed{\textcolor{red}{(x^2=t)}}\\ t^2-5t+4=0 \\ \downarrow\\ (t-4)(t-1)=0 \\ \downarrow\\ t-4=0\rightarrow\boxed{t=4}\\ t-1=0\rightarrow\boxed{t=1}\\ \boxed{t=1,4}\\ \updownarrow\updownarrow\\ \boxed{x^2=t}\\ \downarrow\\ t=1\textcolor{blue}{\rightarrow} x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm1}\\ t=4\textcolor{blue}{\rightarrow} x^2=4\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\textcolor{blue}{\rightarrow} \boxed{x=\pm2}\\ \downarrow\\ \boxed{x=\pm1,\pm2}$

Therefore the given equation has 4 different solutions,

Which means the correct answer is answer D.