The equation x⁴ − 5x² + 4 = 0 has a fourth power, so it isn't a quadratic. But look more closely: x⁴ = (x²)². If you rename x² as t for a moment, you get t² − 5t + 4 = 0, an ordinary quadratic equation.
This trick is called substitution. Together with equations that have the unknown in the denominator, it is one of the most common ways a harder equation turns into a quadratic.
An equation of the form ax⁴ + bx² + c = 0 is a quadratic in disguise (it is sometimes called biquadratic). Here's how to solve it:
Let t = x². The equation becomes at² + bt + c = 0.
Solve it for t.
Go back to x: each positive t gives two solutions x = ±√t. Throw away any negative t, because x² can't be negative.
Worked example 1
x⁴ − 5x² + 4 = 0
t = x²: t² − 5t + 4 = 0.
t₁ = 1, t₂ = 4 (they multiply to 4 and add to 5).
x² = 1 gives x = ±1, x² = 4 gives x = ±2.
Solutions: x = −2, −1, 1, 2. Check for x = 2: 16 − 20 + 4 = 0 ✓.
Worked example 2
x⁴ − 8x² − 9 = 0
t² − 8t − 9 = 0, t₁ = 9, t₂ = −1.
x² = 9 gives x = ±3.
x² = −1 has no solution.
Solutions: x = −3, 3.
Worked example 3
x⁴ + 7x² + 12 = 0
t² + 7t + 12 = 0 has roots t = −3 and t = −4. Both are negative, so the equation has no real solutions. (You can see this without any working too: x⁴ and 7x² are never negative, so the left-hand side is at least 12.)
Substituting a whole expression
If a whole expression is repeated in the equation, you can replace it with a single letter.
Solutions: x = −1, 1, 2, 4. Check for x = 1: (1 − 3)² − 2 · (−2) − 8 = 4 + 4 − 8 = 0 ✓.
The unknown in the denominator
If x is in the denominator, first write down the conditions: a denominator must not be zero. Then multiply the whole equation by the denominator (or by the product of the denominators).
Worked example 5
x − 12/x = 1
Condition: x ≠ 0.
Multiply by x: x² − 12 = x.
Rearrange: x² − x − 12 = 0, roots x = 4 and x = −3.
