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1st BTL CCSS · Equations and systems Notes 22 Oct 2025

1. Biquadratic equations

The substitution $x^{n}=z$ that turns a degree-$2n$ equation into a quadratic, worked examples and 4 solved exercises with hidden solutions.

What is a biquadratic equation?

Definition

A biquadratic equation is an equation that can be turned into a quadratic equation through a change of variable.

The typical case is an even-degree equation in which only the leading term, one middle-degree term and the constant term appear — and the exponents are such that one is double the other. For instance, the highest degree is $2n$ and the middle one is $n$:

$$a\,x^{2n} + b\,x^{n} + c = 0.$$

With the substitution $\;x^{n} = z\;$ we get:

$$a\,z^{2} + b\,z + c = 0,$$

a quadratic equation in $z$, which we already know how to solve.

General strategy

  1. Spot the substitution: if the equation has terms $x^{2n}, x^{n}$ and a constant, set $\;z = x^{n}$.
  2. Rewrite the equation in $z$: you get a quadratic equation.
  3. Solve the quadratic with the formula (or by factoring): you obtain one or two values of $z$.
  4. For each value of $z$, undo the substitution ($x^{n} = z$) and find the roots in $x$.
  5. Collect all the solutions.

Watch out for extra solutions (and the ones you lose)

If you get $\,z<0\,$ and $n$ is even (for example $x^{2}=z$), that value gives no real root, because no real square is negative.

On the other hand, if $n$ is odd ($x^{3}=z$), any value of $z$ gives a single real root, even a negative one: $x = \sqrt[3]{z}$.

Worked examples

Example 1 — just the substitution

Consider $\;5x^{4} - 3x^{2} + 8 = 0$. We make the substitution $\;x^{2} = z$:

$$5z^{2} - 3z + 8 = 0.$$

The discriminant is $\Delta = 9 - 4\cdot 5\cdot 8 = -151 < 0$, so the quadratic has no real solutions. Therefore the original equation has none either.

Example 2 — $2x^{4} - 26x^{2} + 72 = 0$

We apply the substitution $\;x^{2}=z$:

$$2z^{2} - 26z + 72 = 0.$$

We solve with the quadratic formula:

$$z = \dfrac{26 \pm \sqrt{676 - 4\cdot 2\cdot 72}}{4} = \dfrac{26 \pm \sqrt{100}}{4} = \dfrac{26 \pm 10}{4}.$$

So,  $z_{1} = \dfrac{36}{4} = 9$  and  $z_{2} = \dfrac{16}{4} = 4$. Now we undo the substitution:

  • If $z = 9$:  $x^{2} = 9 \;\Longrightarrow\; x = \pm\sqrt{9} = \pm 3$.
  • If $z = 4$:  $x^{2} = 4 \;\Longrightarrow\; x = \pm\sqrt{4} = \pm 2$.
$$\boxed{\;x \in \{-3,\,-2,\,2,\,3\}.\;}$$

A biquadratic equation with the substitution $x^{2}=z$ can have up to four real solutions (two for each $z>0$).

Example 3 — cubic substitution $x^{6} - 9x^{3} + 8 = 0$

Here the exponents are $6$ and $3$, so the substitution is $\;x^{3} = z$:

$$z^{2} - 9z + 8 = 0 \;\Longrightarrow\; z = \dfrac{9 \pm \sqrt{81-32}}{2} = \dfrac{9 \pm 7}{2}.$$

So,  $z_{1} = 8$  and  $z_{2} = 1$. We undo the substitution (now with a cube root):

  • If $z = 8$:  $x^{3} = 8 \;\Longrightarrow\; x = \sqrt[3]{8} = 2$.
  • If $z = 1$:  $x^{3} = 1 \;\Longrightarrow\; x = \sqrt[3]{1} = 1$.
$$\boxed{\;x \in \{1,\,2\}.\;}$$

With an odd substitution ($x^{3}=z$) each $z$ gives a single real root, so in this case we have at most two solutions.

Exercises

30 Solve the biquadratic equations

Solve the proposed biquadratic equations. Then check the results you obtained with GeoGebra.

a) $x^{4} - 5x^{2} + 4 = 0$

Substitution $\;x^{2} = t$:

$$t^{2} - 5t + 4 = 0 \;\Longrightarrow\; t = \dfrac{5 \pm \sqrt{25-16}}{2} = \dfrac{5 \pm 3}{2}.$$

$t_{1} = 4$  and  $t_{2} = 1$. We undo the substitution:

  • If $x^{2} = 1$:  $x = \pm 1$.
  • If $x^{2} = 4$:  $x = \pm 2$.
$$\boxed{\;x \in \{-2,\,-1,\,1,\,2\}.\;}$$
b) $x^{4} - 2x^{2} + 1 = 0$

Substitution $\;x^{2} = t$:

$$t^{2} - 2t + 1 = 0 \;\Longrightarrow\; (t-1)^{2} = 0 \;\Longrightarrow\; t = 1\ \text{(double)}.$$

We undo the substitution: if $x^{2} = 1$, then $x = \pm 1$. Since $t=1$ was a double root, each of these is double:

$$\boxed{\;x = 1\ \text{(double)},\quad x = -1\ \text{(double)}.\;}$$

It is clearer by factoring: $\;x^{4} - 2x^{2} + 1 = (x^{2}-1)^{2} = (x-1)^{2}(x+1)^{2}$.

c) $x^{4} - 3x^{2} = 0$

Here no change of variable is needed: we factor out $x^{2}$ and get a product that is zero whenever one of its factors is:

$$x^{2}(x^{2} - 3) = 0.$$
  • $x^{2} = 0$:  $x = 0\ \text{(double)}$.
  • $x^{2} - 3 = 0$:  $x = \pm\sqrt{3}$.
$$\boxed{\;x = 0\ \text{(double)},\quad x = \pm\sqrt{3}.\;}$$

Factored: $\;x^{4} - 3x^{2} = x^{2}\,(x-\sqrt{3})(x+\sqrt{3})$.

d) $x^{6} - 7x^{3} - 8 = 0$

The exponents are $6$ and $3$, so the substitution is $\;x^{3} = t$:

$$t^{2} - 7t - 8 = 0 \;\Longrightarrow\; t = \dfrac{7 \pm \sqrt{49+32}}{2} = \dfrac{7 \pm 9}{2}.$$

$t_{1} = 8$  and  $t_{2} = -1$. We undo the substitution (cube root):

  • If $x^{3} = 8$:  $x = \sqrt[3]{8} = 2$.
  • If $x^{3} = -1$:  $x = \sqrt[3]{-1} = -1$.
$$\boxed{\;x \in \{-1,\,2\}.\;}$$

The cube root of a negative number exists in the reals, so the case $t=-1$ is not discarded — that only happens with the substitution $x^{2}=z$.