Substitution

In one sentence

x^{4} - 5 x^{2} + 4 = (x^{2})^{2} - 5 (x^{2}) + 4 t = x^{2} t^{2} - 5 t + 4

An expression is often long not because it is truly complicated, but because the same block has been written out several times. Give that block a name, $t$ , and the expression shrinks into a small problem you already know how to solve. Finish the job in the $t$ -world, then switch $t$ back to the block it stood for.

Substitution has three steps: find the repeated block → solve in the new world → substitute back. The one most often forgotten is the third.

Example: solve x⁴ − 5x² + 4 = 0

Step 1 / 4

Find the repeated block
$x^{4}$ is just $(x^{2})^{2}$ . Throughout the equation, $x$ only ever appears as $x^{2}$ .

$(x^{2})^{2} - 5 x^{2} + 4 = 0$
Put the lens on
Solve in the t-world
Take the lens off

Predict first

Use the same method on $x^{4} - 5 x^{2} - 36 = 0$ . How many real solutions does it have?

4

2

0

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

Stop 1 · Factorization (Vol. 1)

The repeated block does not have to be $x^{2}$ ; it can be a whole quadratic.

Example: factor (x² + 3x)² − 2(x² + 3x) − 8

Step 1 / 4

Do not expand yet
Expanding gives a quartic (the top line in the figure), which hides what is going on. Keep the structure: $x^{2} + 3 x$ appears twice.
Substitute
Cross method
Substitute back, then check

Stop 2 · Higher-degree equations (Vol. 2)

Sometimes the repeated block is hidden, and you have to build it yourself first.

Example: solve (x+1)(x+2)(x+3)(x+4) = 24

Step 1 / 5

How to pair them
Multiplying out all four brackets gives a quartic equation. Change the order of multiplication instead: the outer two as one pair, the inner two as another. Since $1 + 4 = 2 + 3$ , the first two terms of the two products are exactly the same.

$x^{2} + 5 x + 4 (x + 1) (x + 4) \cdot x^{2} + 5 x + 6 (x + 2) (x + 3) = 24$
Substitute
Solve for t
Substitute back
Check

Stop 3 · Fractional equations (Vol. 2)

When $x$ and $\frac{1}{x}$ both appear, keep your eye on $x + \frac{1}{x}$ . Its square has $x^{2} + \frac{1}{x ^{2}}$ hidden inside.

Example: solve x² + 1/x² − 3(x + 1/x) + 4 = 0

Step 1 / 6

Find the link
Square $x + \frac{1}{x}$ : the middle term $2 \cdot x \cdot \frac{1}{x}$ is exactly the constant $2$ .

$(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x ^{2}} \Rightarrow x^{2} + \frac{1}{x ^{2}} = t^{2} - 2$
Substitute
Solve for t
Substitute back t = 2
Substitute back t = 1
Conclusion

Stop 4 · Radical equations (Vol. 2)

For an equation with a square root, squaring is one way in, but squaring brings in fake solutions. Substitution is cleaner: call the whole square root $t$ , and write down $t \geq 0$ .

Example: solve x − 3 = √(x + 3)

Step 1 / 5

Make x + 3 appear outside the root too
The left side is $x - 3 = (x + 3) - 6$ . After moving terms across, $x + 3$ appears once inside the root and once outside.

$(x + 3) - x + 3 - 6 = 0$
Substitute, and bring the range along
Solve for t
Substitute back
Compare with “square both sides”

Practice

Exercise 1Basic

Solve $x^{4} - 13 x^{2} + 36 = 0$ . (Enter all the real solutions, separated by commas.)

Exercise 2Advanced

Factor: $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12$ .

Exercise 3Challenge

Solve $x (x + 1) (x + 2) (x + 3) = 24$ . (Enter all the real solutions, separated by commas.)

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Forgetting that the new variable has its own range

Solve

x^{4} - 5 x^{2} - 36 = 0

: let

t = x^{2}

, so

t = 9

t = - 4

. Hence

x = \pm 3

x = \pm 2

What went wrong? Think first, then open

Diagnosis $t$ is not just any number: it is $x^{2}$ , so $t \geq 0$ . $t = - 4$ would mean $x^{2} = - 4$ , and no real $x$ can do that. Treating $- 4$ as if it were $4$ and taking square roots creates two fake solutions out of nowhere.

Reject

t = - 4

. Only

x^{2} = 9

is left, so

x = \pm 3

. Substitute back into the original equation:

81 - 45 - 36 = 0

. It works.

“When you substitute, carry the new variable's range across too”

Exhibit

Staying in the t-world and never coming back

Factor

(x^{2} + 3 x)^{2} - 2 (x^{2} + 3 x) - 8

: let

t = x^{2} + 3 x

, giving

(t - 4) (t + 2)

. So the answer is

(x^{2} + 3 x - 4) (x^{2} + 3 x + 2)

What went wrong? Think first, then open

Diagnosis Only half of the substituting back was done. Once the lens comes off, each factor must be checked again in the $x$ -world: $x^{2} + 3 x - 4$ and $x^{2} + 3 x + 2$ can both still be factored by the cross method.

(x^{2} + 3 x - 4) (x^{2} + 3 x + 2) = (x + 4) (x - 1) (x + 1) (x + 2)

“After substituting back, ask once more “can it factor further?””

In one sentence

Stop 1 · Factorization (Vol. 1)

Stop 2 · Higher-degree equations (Vol. 2)

Stop 3 · Fractional equations (Vol. 2)

Stop 4 · Radical equations (Vol. 2)

Practice

Mistake Museum · Exhibits from this stop

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