Chapter 1.6, Problem 17E

To determine

To calculate: The solution of the equation $x^4-4x^2+3=0$.

The solution of the equation is $x=\pm \sqrt{3},\pm 1$.

Calculation:

Consider the provided equation,

$x^4-4x^2+3=0$

Put $x^2=u$ and convert this equation into the quadratic form,

$u^2-4u+3=0$

Now, factorize the above equation,

$\left(u-3\right)\left(u-1\right)=0$

Put the first factor equal to zero, then replace $u$ with $x^2$ and solve for $x$,

$\begin{array}{c}u-3=0\\ u=3\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

Put the second factor equal to zero, then replace $u$ with $x^2$ and solve for $x$,

$\begin{array}{c}u-1=0\\ u=1\\ x^2=1\\ x=\pm 1\end{array}$

Check:

Put $x=\pm \sqrt{3},\pm 1$ in the equation,

First, put $x=\sqrt{3}$,

$\begin{array}{r}{\left(\sqrt{3}\right)}^4-4{\left(\sqrt{3}\right)}^2+3\stackrel{?}{=}0\\ 9-12+3\stackrel{?}{=}0\\ 12-12\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Now, put $x=-\sqrt{3}$,

$\begin{array}{r}{\left(-\sqrt{3}\right)}^4-4{\left(-\sqrt{3}\right)}^2+3\stackrel{?}{=}0\\ 9-12+3\stackrel{?}{=}0\\ 12-12\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Now, put $x=1$,

$\begin{array}{r}{\left(1\right)}^4-4{\left(1\right)}^2+3\stackrel{?}{=}0\\ 1-4+3\stackrel{?}{=}0\\ 4-4\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Now, put $x=-1$,

$\begin{array}{r}{\left(-1\right)}^4-4{\left(-1\right)}^2+3\stackrel{?}{=}0\\ 1-4+3\stackrel{?}{=}0\\ 4-4\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Hence, the solution of equation is $x=\pm \sqrt{3},\pm 1$.