What is a hidden quadratic equation?

An equation of the form a[f(x)]² + b[f(x)] + c = 0, where x has been replaced by a function f(x). It becomes a standard quadratic when we substitute u = f(x).

What substitution converts x⁴ - 5x² + 4 = 0 to a quadratic?

Let u = x². The equation becomes u² - 5u + 4 = 0.

For sin²x + 2sin x - 3 = 0, what is the corresponding normal quadratic?

x² + 2x - 3 = 0, where f(x) = sin x. The structure is identical; only the variable changes.

Why might sin x = -3 have no solution?

The range of sin x is -1 ≤ sin x ≤ 1. Since -3 is outside this range, no real angle satisfies sin x = -3.

What are the four steps to solve a hidden quadratic?

1) Rearrange to a[f(x)]² + b[f(x)] + c = 0; 2) Substitute u = f(x); 3) Solve au² + bu + c = 0 for u; 4) Substitute back and solve f(x) = u for x.

Solve x⁴ - 13x² + 36 = 0 using substitution.

Let u = x²: u² - 13u + 36 = 0. Factor: (u-4)(u-9)=0. So u=4 or u=9. Thus x²=4 → x=±2, and x²=9 → x=±3. Solutions: x = ±2, ±3.

What substitution would you use for 2e²ˣ - 5eˣ + 2 = 0?

Let u = eˣ. The equation becomes 2u² - 5u + 2 = 0.

If u = x² and u = 9, what are the possible values of x?

x² = 9 gives x = 3 or x = -3. Always consider both positive and negative roots.

What is the quadratic formula?

x = (-b ± √(b² - 4ac)) / (2a) for ax² + bx + c = 0. Use this when factorisation is difficult.

How do you recognise a hidden quadratic in trigonometry?

Look for equations with both sin²x (or cos²x) and sin x (or cos x) terms, plus a constant. The squared and linear terms in the same function indicate a hidden quadratic.

Solving Hidden Quadratics using Substitutions | Pearson Edexcel International AS Maths

Solving Hidden Quadratics using Substitutions

Summary

Hidden quadratic equations are quadratics written in terms of a function f(x) rather than x. By substituting u = f(x), we can reduce them to standard quadratics, solve for u, then substitute back to find x.

1. What Are Hidden Quadratic Equations?

A normal quadratic appears in the form (ax^2 + bx + c = 0).

A hidden quadratic appears in the form (a[f(x)]^2 + b[f(x)] + c = 0), where (x) has been replaced by some function (f(x)).

Example: (\sin^2 x + 2\sin x - 3 = 0) is the hidden quadratic of (x^2 + 2x - 3 = 0) where (f(x) = \sin x).

2. Recognising Hidden Quadratics

Look for equations where:

A squared term appears: ([f(x)]^2)
A linear term in the same function: (f(x))
A constant term

Common functions include: (\sin x), (\cos x), (e^x), (2^x), or (x^2) (giving (x^4) terms).

3. Solution Method

Step 1: Rearrange into the form (a[f(x)]^2 + b[f(x)] + c = 0)

Step 2: Substitute (u = f(x)) to get (au^2 + bu + c = 0)

Step 3: Solve the standard quadratic for (u) (factorise, complete the square, or use the quadratic formula)

Step 4: Substitute back: set (f(x) = u) for each solution and solve for (x)

4. Worked Example

Solve (2x^4 - 7x^2 + 3 = 0)

Let (u = x^2). Then (2u^2 - 7u + 3 = 0)

Factorising: ((2u - 1)(u - 3) = 0)

So (u = \frac{1}{2}) or (u = 3)

Substituting back: (x^2 = \frac{1}{2}) gives (x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2})

(x^2 = 3) gives (x = \pm \sqrt{3})

Solutions: (x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \sqrt{3}, -\sqrt{3})

5. Trigonometric Hidden Quadratics

6. Common Pitfalls