Quick questions on Equations reducible to quadratic form explained: O-Level A-Maths

An equation is reducible to quadratic form when one power or expression is the square of another. Tell-tale signs:

The fastest way to choose the substitution is to look for a term that is the square of another term in the equation. Whenever you see a power that is exactly double another, $x^4$ doubling $x^2$, $a^{2x}$ doubling $a^x$, or $x$ as the square of $\sqrt{x}$, let $u$ be the smaller of the pair. The equation then collapses to $au^2 + bu + c = 0$. Checking that the highest power is precisely twice the middle power before substituting confirms the equation really is reducible; if the powers are not in a $2:1$ ratio, no single substitution will make it quadratic and a different method is needed.

Knowing how many solutions to expect guards against losing some. A quadratic in $u$ gives up to two values of $u$, and each is reverted to the original variable, so the final solution count depends on the substitution: $u = x^2$ can give up to four real values of $x$ (two signs for each positive $u$), while $u = a^x$ gives at most one $x$ per valid positive $u$. So $x^4 - 13x^2 + 36 = 0$ has four roots, but $3^{2x} - 4(3^x) + 3 = 0$ has only two. Predicting the expected number of roots from the substitution is a built-in check that you have not dropped a sign or an impossible value.

Substitution can create spurious roots; verify each reverted value satisfies the original equation.

Solve $x^4 - 5x^2 + 4 = 0$. [3 marks]

Solve $2^{2x} - 6(2^x) + 8 = 0$. [4 marks]

Solve $x - 4\sqrt{x} + 3 = 0$. [3 marks]