Higher-Degree Equations

What this move does

Cubic and quartic equations have no handy all-purpose formula. Our plan fits in three words: reduce the degree. Turn the equation into linear and quadratic equations. There are two common routes:

Substitution: if a “block” appears again and again (such as $x^{2}$ , $x^{2} + x$ or $x + \frac{1}{x}$ ), treat that block as a new letter $t$ . The higher-degree equation becomes a quadratic in $t$ .
Solve for $t$ , then send it back: each value of $t$ gives a lower-degree equation in $x$ . Some values of $t$ cannot be sent back (for example $x^{2} = - 2$ ), and that branch has no real roots.
Factorization: if the left side factors into lower-degree factors and the right side is $0$ , set each factor equal to $0$ in turn.

The simplest example is a biquadratic equation, with only $x^{4}$ , $x^{2}$ and a constant: $x^{4} - 5 x^{2} + 4 = 0$ . Cover up $x^{2}$ and call it $t$ , and it becomes $t^{2} - 5 t + 4 = 0$ .

Predict first

How many real roots does $x^{4} - x^{2} - 6 = 0$ have? (Hint: with $t = x^{2}$ it becomes $t^{2} - t - 6 = 0$ .)

4

2

0

3

Choose an answer first; then you can see the result.

Work through one

The block does not have to be $x^{2}$ . Expanded, the equation below is $x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 = 0$ , which looks scary. But this is how it was actually written:

Example: solve (x² + x)² − 8(x² + x) + 12 = 0

Step 1 / 5

Find the repeated block
$x^{2} + x$ appears twice: once squared, once not. Whatever you do, do not rush to expand: that just gives the quartic above.

$(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0$
Put the lens on
Send t = 2 back
Send t = 6 back
Put them together and check the graph

Equations with symmetric coefficients

The coefficients of $2 x^{4} - 9 x^{3} + 14 x^{2} - 9 x + 2 = 0$ are $2, - 9, 14, - 9, 2$ : the same read from the left or from the right. Equations like this have a ready-made block: $x + \frac{1}{x}$ .

Since $x = 0$ is not a root, we can divide both sides by $x^{2}$ and pair the first and last terms:

2 (x^{2} + \frac{1}{x ^{2}}) - 9 (x + \frac{1}{x}) + 14 = 0.

Let $t = x + \frac{1}{x}$ . Then $x^{2} + \frac{1}{x ^{2}} = t^{2} - 2$ , and the equation becomes $2 (t^{2} - 2) - 9 t + 14 = 0$ , that is, $2 t^{2} - 9 t + 10 = 0$ .

Put the lens on, then click each

t

in turn. To send

t

back, multiply both sides of

x + \frac{1}{x} = t

x

: you get

x^{2} - t x + 1 = 0

Notice that the branch $t = 2$ gives only one value, $x = 1$ (a double root), so the curve just touches the $x$ -axis at $x = 1$ .

The other route: factorization

When there is no obvious block, try the method from Vol. 1: guess a rational root, then divide out the matching linear factor.

Example: solve x³ − 2x² − 5x + 6 = 0

Step 1 / 4

Try some roots
The leading coefficient is $1$ , so any rational root must be a divisor of the constant $6$ : $\pm 1, \pm 2, \pm 3, \pm 6$ . Try $x = 1$ : $1 - 2 - 5 + 6 = 0$ . A hit.

$f (1) = 1 - 2 - 5 + 6 = 0$
Divide out the linear factor
What is left is a quadratic
Write down all the roots

Practice

Exercise 1Basic

Solve $x^{4} - 10 x^{2} + 9 = 0$ . (Give all the real roots, separated by commas.)

Exercise 2Basic

Solve $x^{3} = 9 x$ .

Exercise 3Advanced

Solve $(x^{2} - 2 x)^{2} - 2 (x^{2} - 2 x) - 3 = 0$ .

Exercise 4Challenge

Solve $x (x + 1) (x + 2) (x + 3) = 24$ .

What this move does

Work through one

Example: solve (x² + x)² − 8(x² + x) + 12 = 0

Equations with symmetric coefficients

The other route: factorization

Example: solve x³ − 2x² − 5x + 6 = 0

Practice

Mistake Museum · Exhibits from this stop

Stopping once you have t

Taking the square root of a negative t

Dividing both sides by x and losing x = 0