Equations Reducible to Quadratic Equations
Many equations that are not quadratic at first sight can be turned into quadratic equations by a substitution or by clearing fractions or roots. In brief: the equation x4 − 5x2 + 4 = 0 becomes u2 − 5u + 4 = 0 with u = x2, so u = 1 or 4 and x = ±1 or ±2. Equations with fractions and roots need a check for extraneous solutions.
Recognising a hidden quadratic form saves a lot of work. The main types are biquadratic equations, equations with an expression repeated in two powers, equations with fractions, equations with radicals and exponential equations. This guide explains each type with a worked example and a table of substitutions.
How do you solve biquadratic equations?
A biquadratic equation has the form ax4 + bx2 + c = 0. Let u = x2 with u ≥ 0. Then au2 + bu + c = 0. Solve for u, discard negative values and then solve x2 = u for each remaining u.
Example 1: x4 − 5x2 + 4 = 0
With u = x2: u2 − 5u + 4 = 0, so (u − 1)(u − 4) = 0 and u = 1 or 4. Then x2 = 1 gives x = ±1 and x2 = 4 gives x = ±2. Four solutions: −2, −1, 1, 2.
Example 2: x4 + 3x2 − 10 = 0
u2 + 3u − 10 = 0 gives (u + 5)(u − 2) = 0, so u = −5 or u = 2. The value u = −5 is rejected because x2 cannot be negative (in the real numbers). From u = 2: x = ±√2.
How do you use other substitutions?
The same idea works whenever an expression and its square appear. In (x2 − 3x)2 − 2(x2 − 3x) − 8 = 0, let u = x2 − 3x. Then u2 − 2u − 8 = 0, so (u − 4)(u + 2) = 0 and u = 4 or u = −2. Back-substitute: x2 − 3x − 4 = 0 gives x = 4 or −1, and x2 − 3x + 2 = 0 gives x = 1 or 2. The four solutions are −1, 1, 2 and 4.
Example 3: powers x and √x
x − 5√x + 6 = 0. Let u = √x ≥ 0. Then u2 − 5u + 6 = 0, so u = 2 or 3, and x = 4 or 9. Check: 4 − 10 + 6 = 0 ✓ and 9 − 15 + 6 = 0 ✓.
Example 4: reciprocals
x + 1/x = 5/2. Multiply by 2x: 2x2 + 2 = 5x, so 2x2 − 5x + 2 = 0 and (2x − 1)(x − 2) = 0. The solutions are x = 1/2 and x = 2 (both nonzero).
How do you solve equations with fractions?
- State the values excluded by the denominators.
- Multiply by the common denominator.
- Solve the resulting polynomial equation.
- Reject excluded values.
Example 5: 2/(x − 1) + 3/(x + 2) = 1
Excluded: x ≠ 1, −2. Multiply by (x − 1)(x + 2): 2(x + 2) + 3(x − 1) = (x − 1)(x + 2). So 5x + 1 = x2 + x − 2, giving x2 − 4x − 3 = 0. Then x = 2 ± √7. Both are allowed. Numerically 4.65 and −0.65. Check x ≈ 4.65: 2/3.65 + 3/6.65 ≈ 0.548 + 0.451 = 0.999 ✓.
How do you solve equations with square roots?
Isolate the root and square both sides, then check, because squaring can add solutions.
Example 6: √(x + 6) = x
Square: x + 6 = x2, so x2 − x − 6 = 0, and (x − 3)(x + 2) = 0. Check x = 3: √9 = 3 ✓. Check x = −2: √4 = 2 ≠ −2. So x = −2 is extraneous, and the only solution is x = 3.
How do you solve exponential equations of quadratic type?
Let u = 2x in 4x − 6 · 2x + 8 = 0. Since 4x = (2x)2, we get u2 − 6u + 8 = 0, so u = 2 or 4. Then 2x = 2 gives x = 1 and 2x = 4 gives x = 2. Because 2x is always positive, any negative u must be rejected.
Table of substitutions
| Equation | Substitution | Solutions |
|---|---|---|
| x4 − 5x2 + 4 = 0 | u = x2 | −2, −1, 1, 2 |
| x − 5√x + 6 = 0 | u = √x | 4, 9 |
| x + 1/x = 5/2 | multiply by x | 1/2, 2 |
| 4x − 6 · 2x + 8 = 0 | u = 2x | 1, 2 |
| (x2 − 3x)2 − 2(x2 − 3x) − 8 = 0 | u = x2 − 3x | −1, 1, 2, 4 |
Where does the technique come from?
Substitution to reduce the degree of an equation appears in Arabic algebra and in European texts of the Renaissance. Biquadratic equations were known to Cardano and his contemporaries in connection with quartic equations. The systematic use of a new variable such as u is part of the symbolic notation developed from the sixteenth century on.
Where is it used today?
Quartic equations of biquadratic type appear in vibration problems and in beam bending, where the unknown is a squared frequency. Equations with √x arise in physics when the speed depends on the root of a quantity. Exponential equations of quadratic type occur in chemistry and in finance when two growth terms interact.
Which mistakes are common?
Stopping at u
After finding u, return to x. The answer to the original problem is x, not u.
Keeping impossible values of u
If u = x2 or u = 2x, then u must be positive (or nonnegative). Reject other values.
Forgetting to check
With fractions and roots, test every solution in the original equation.
Losing solutions
Each positive u = x2 gives two values of x, ±√u.