As in the previous example, we note that after a substitution of [latex]u=x^2[/latex], the equation becomes quadratic. We can solve the resulting quadratic equation by factoring (or quadratic formula if necessary).

[latex]\begin{align}x^4-3x^2-4&=0\\ (x^2)^2-3(x^2)-4&=0\\ u^2-3u-4&=0\\ (u-4)(u+1)&=0\\ u-4=0 \textsf{ or } u+1=0 &\\ u=4 \textsf{ or } u=1& \end{align}[/latex]

Now substitute [latex]x^2[/latex] for [latex]u[/latex] to solve for the original variable in each equation.

[latex]\begin{align}x^2&=4\\ x&=\pm \sqrt{4}\\ x&=2, -2 \end{align}[/latex]

[latex]\begin{align}x^2&=-1\\ x&=\pm \sqrt{-1}\\ x&=i, -i \end{align}[/latex]

The four complex solutions are [latex]x=2, -2, i, -i[/latex].

If the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[/latex].

This suggests using a substitution [latex]u=x^{-1}[/latex]. We verify that that [latex]u^2 = (x^{-1})^2 = x^{-2}[/latex], which allows us to substitute the first term as well. Now we obtain

[latex]\begin{align}x^{-2}+5x^{-1}+6&=0\\ (x^{-1})^2+5(x^{-1})+6&=0\\ u^2+5u+6&=0\\ (u+2)(u+3)&=0\\ u+2=0 \textsf{ or } u+3=0 &\\ u=-2 \textsf{ or } u=-3& \end{align}[/latex]

Now substitute [latex]x^{-1}[/latex] for [latex]u[/latex] to solve for the original variable in each equation.

[latex]\begin{align}x^{-1}&=-2\\ \dfrac{1}{x}&=-2\\ 1&=-2x\\ x&=-\dfrac{1}{2} \end{align}[/latex]

[latex]\begin{align}x^{-1}&=-3\\ \dfrac{1}{x}&=-3\\ 1&=-3x\\ x&=-\dfrac{1}{3} \end{align}[/latex]

The two solutions are [latex]x=-\dfrac{1}{2}[/latex] or [latex]x=-\dfrac{1}{3}[/latex].

We begin by getting everything on the left side. Since a square root is equivalent to an exponent of [latex]\dfrac{1}{2}[/latex], this suggests a “descending order” to place the terms in with the square root as the middle term.

[latex]\begin{align}\sqrt{x}-x&=-2\\ \sqrt{x}-x+2&=0\\ -x+\sqrt{x}+2&=0 \end{align}[/latex]

The middle term suggests using the substitution [latex]u=\sqrt{x}[/latex]. We verify that [latex]u^2=(\sqrt{x})^2=x[/latex], which replaces the first term. Thus we continue

[latex]\begin{align} -u^2+u+2&=0\\ -(u^2-u-2)&=0\\ -(u-2)(u+1)&=0\\ u-2=0 \textsf{ or } u+1=0 &\\ u=2 \textsf{ or } u=-1& \end{align}[/latex]

Now substitute [latex]\sqrt{x}[/latex] for [latex]u[/latex] to solve for the original variable in each equation.

[latex]\begin{align}\sqrt{x}&=2\\ (\sqrt{x})^2&=2^2\\ x&=4 \end{align}[/latex]

[latex]\sqrt{x}=-1[/latex]

The second equation cannot be solved, as the principal square root is always positive. Thus the solution is [latex]x=4[/latex].