Spot the pattern, rename it, solve the quadratic, then change back.
Solve for x
x⁴ − 13x² + 36 = 0
Let u = x². Then u² − 13u + 36 = 0 factors as (u − 4)(u − 9) = 0, so u = 4 or 9. Changing back: x² = 4 gives x = ±2, and x² = 9 gives x = ±3 — four solutions. A stops at u and never converts back. B forgets the negative roots — a fourth-degree equation can have four.
The middle exponent is half the first, and there is a constant.
| Equation | Substitute | Becomes |
|---|---|---|
| x⁴ − 13x² + 36 = 0 | u = x² | u² − 13u + 36 = 0 |
| x⁶ − 7x³ − 8 = 0 | u = x³ | u² − 7u − 8 = 0 |
| x − 5√x + 6 = 0 | u = √x | u² − 5u + 6 = 0 |
| (x+1)² − 4(x+1) + 3 = 0 | u = x + 1 | u² − 4u + 3 = 0 |
Always check the exponent pattern. x⁵ − 3x² + 1 is not in quadratic form, because 2 is not half of 5, and no substitution turns it into one.
x⁴ − 13x² + 36 = 0
let u = x²
u² − 13u + 36 = 0
an ordinary quadratic
(u − 4)(u − 9) = 0
u = 4 or 9
x² = 4 → x = ±2
change back
x² = 9 → x = ±3
four solutions in all
Stopping at u = 4 answers a question nobody asked. The substitution is a device for making the equation solvable; the last step always undoes it.
x⁴ + 3x² − 4 = 0
u = x²
(u + 4)(u − 1) = 0
u = −4 or 1
x² = 1 → x = ±1
fine
x² = −4 → no real x
rejected
Two real solutions, not four. A negative value of u is perfectly legitimate as a value of u — it just produces no real x when you convert back.
xx − 2 + 3 = 6x − 2
x ≠ 2
x + 3(x − 2) = 6
× (x − 2)
4x − 6 = 6, x = 3
3 ≠ 2, so it stands ✓
Whatever makes a denominator zero can never be a solution, however tidily the algebra produces it. Write those values down while the denominators are still visible.
x − 5√x + 6 = 0
let u = √x
u² − 5u + 6 = 0
since (√x)² = x
u = 2 or u = 3
both positive, both usable
x = 4 or x = 9
square each
Whatever u stands for, changing back means undoing exactly that.
u = x³: solve x⁶ − 26x³ − 27 = 0
u² − 26u − 27 = 0
u = x³
(u − 27)(u + 1) = 0
u = 27 or u = −1
x³ = 27 → x = 3; x³ = −1 → x = −1
take cube roots
x = 3 or x = −1
check: 729 − 702 − 27 = 0 ✓
Copying the ± habit from x² gives x = ±3 and x = ±1. But (−3)³ = −27, not 27, and 1³ = 1, not −1. Only x² = k gives a ± pair; x³ = k gives one real answer, with the same sign as k.
u = 1x: solve 3x² − 5x − 2 = 0
3u² − 5u − 2 = 0
1x² = u²
(3u + 1)(u − 2) = 0
u = −13 or u = 2
x = −3 or x = 12
x is the reciprocal of u
Check x = 12: 3 ÷ 14 − 5 ÷ 12 − 2 = 12 − 10 − 2 = 0 ✓
u = a bracket: solve (3x + 1)² − 9(3x + 1) + 20 = 0
u² − 9u + 20 = 0
u = 3x + 1; do not expand
(u − 4)(u − 5) = 0
u = 4 or u = 5
3x + 1 = 4 or 3x + 1 = 5
change back
x = 1 or x = 43
check x = 1: 16 − 36 + 20 = 0 ✓
u = x1/3: solve x2/3 − 3x1/3 − 4 = 0
u² − 3u − 4 = 0
x2/3 = (x1/3)² = u²
(u − 4)(u + 1) = 0
u = 4 or u = −1
3√x = 4 → x = 4³ = 64; 3√x = −1 → x = −1
cube each value of u
x = 64 or x = −1
check x = 64: 16 − 12 − 4 = 0 ✓
The same u = x² substitution; what changes is what u turns out to be.
u = 0 is a root: x⁴ − 9x² = 0
u² − 9u = 0, so u(u − 9) = 0
u = 0 or u = 9
x = 0 or x = ±3
x² = 0 gives only 0
Different: three distinct solutions, not four — zero has no ± partner.
u is not a perfect square: x⁴ − 7x² + 10 = 0
(u − 2)(u − 5) = 0
u = 2 or u = 5
x = ±√2 or x = ±√5
check: 4 − 14 + 10 = 0 ✓
Different: the four roots are irrational, but the method is unchanged.
Both values of u are negative: x⁴ + 6x² + 8 = 0
(u + 2)(u + 4) = 0
u = −2 or u = −4
no real solutions
x² cannot be negative
Different: every term on the left is positive, so the left side is at least 8 and never 0.
Expect "which substitution", "how many real solutions" and full solve items for x⁴, √x, x1/3 or a repeated bracket. The fastest route is to name u, solve the quadratic in u, then change back carefully — ± for x², one root for x³, reject negatives for x² and √x. The 10-second check: count the solutions each u gives and substitute one root into the original.
Solve for x
x⁴ − 5x² + 4 = 0
With u = x², the equation becomes u² − 5u + 4 = (u − 1)(u − 4) = 0, so u = 1 or 4. Converting back: x² = 1 gives x = ±1, and x² = 4 gives x = ±2 — four solutions. A leaves the answer in terms of u, and D drops the negative roots.
Hint: a fourth-degree equation can have up to four roots. If you have only two, check whether a ± went missing when you converted back.
