With u = x², the equation gives u = 3. What is x?

√3 or −√3. x² = 3 has two solutions.

With u = eˣ, the equation gives u = −2. What follows?

there is no solution from this value. eˣ is always positive.

How many real solutions does x⁴ − 5x² + 4 = 0 have?

4. x² = 1 and x² = 4 each give two.

Equations that are quadratics in disguise · Cambridge IGCSE Additional Mathematics

In plain words

x⁴ − 5x² + 4 = 0 looks like a new kind of problem. It is not. Call x² by another name, u, and it becomes u² − 5u + 4 = 0: an ordinary quadratic. The trick is spotting what to rename.

5 things to know

Look for one expression and its square: x² and x⁴, eˣ and e^(2x), √x and x, ln x and (ln x)².
Let u stand for the simpler expression. The equation becomes a quadratic in u.
Solve the quadratic for u. Then turn each value of u back into x.
Some values of u give no solution: x² cannot equal a negative number, and neither can eˣ or √x.
If e⁻ˣ appears with eˣ, multiply every term by eˣ first.

Worked example

Solve e^(2x) − 5eˣ + 6 = 0.

e^(2x) is (eˣ)². Let u = eˣ: u² − 5u + 6 = 0.
Factorise: (u − 2)(u − 3) = 0, so u = 2 or u = 3.
eˣ = 2 gives x = ln 2. eˣ = 3 gives x = ln 3.

Worked example

Solve 2eˣ = 7 − 3e⁻ˣ.

Multiply every term by eˣ: 2e^(2x) = 7eˣ − 3. Let u = eˣ: 2u² − 7u + 3 = 0.
Factorise: (2u − 1)(u − 3) = 0, so u = 1/2 or u = 3.
x = ln(1/2), which is −ln 2, or x = ln 3.

Tips and tricks

Do not stop at u. The question asks for x, so every value of u has to be converted back.
State the substitution you are using. It earns the first method mark.

5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Equations that are quadratics in disguise: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

Which substitution turns x⁴ − 7x² + 12 = 0 into a quadratic?
- u = x
- u = x² (the answer)
- u = x⁴
- u = 7x
Then x⁴ is u².
How can e^(2x) be written in terms of u = eˣ?
- 2u
- u² (the answer)
- u + 2
- e^u
(eˣ)² = e^(2x).
With u = x², the equation gives u = 3. What is x?
- 3
- 9
- √3 or −√3 (the answer)
- 1.5
x² = 3 has two solutions.
With u = eˣ, the equation gives u = −2. What follows?
- x = ln 2
- x = −ln 2
- there is no solution from this value (the answer)
- x = −2
eˣ is always positive.
How many real solutions does x⁴ − 5x² + 4 = 0 have?
- 1
- 2
- 3
- 4 (the answer)
x² = 1 and x² = 4 each give two.

5 questions

Tap an answer and you’ll see straight away whether it’s right, and why.

Worksheet

3 questions, 10 marks. Write your answers on paper, then check them.